InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 12 knowledge and support exam preparation. Choose a topic below to get started.
| 301. |
A fringe width of a certain interferences pattern is beta = 0.002 cm. What is distance of 5th dark fringe from centre ? |
| Answer» <html><body><p>`9 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a>`<br/>`11 xx 10^(-2) cm`<br/>`1.1 xx 10^(-2) cm`<br/>`3.28 xx 10^(6) cm`</p>Solution :The distance of 5m <a href="https://interviewquestions.tuteehub.com/tag/dark-943775" style="font-weight:bold;" target="_blank" title="Click to know more about DARK">DARK</a> fringe from centre is given by : <br/> `x_(n) = (<a href="https://interviewquestions.tuteehub.com/tag/2n-300431" style="font-weight:bold;" target="_blank" title="Click to know more about 2N">2N</a> + 1)(lambdal)/(2d)` as n = 4 for 5th dark fringe <br/> So, `x_(5) = (9)/(2) (lambdaD)/(d)` <br/> as `(lambdaD)/(d) = beta` <br/> So, `x_(5) = (9/2)beta = 9/2 xx 0.002` <br/> ` = 9 xx 10^(-3) cm`.</body></html> | |
| 302. |
Monochromatic light of requency 6.0xx10^(14) Hz is produced by a laser. The power emitted is 2.0xx10^(-3)W. Calculate the (i) energy of a photon in the light beam, and (ii) number of photons emitted on an average by the source. |
| Answer» <html><body><p></p>Solution :Here <a href="https://interviewquestions.tuteehub.com/tag/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a> `v=6.0xx10^(<a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a>)Hz` and power of laser `P=2.0xx10^(-3)W` <br/> (i) Energy of a photon `E=hv=6.63xx10^(-34)xx6.0xx10^(14)=3.978xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>)J=3.98xx10^(-19)J` <br/> (ii) Number of photons emitted <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> second by laser source <br/> `n=(P)/(E)=(2.0xx10^(-3))/(3.98xx10^(-19))=5.03xx10^(15)s^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`.</body></html> | |
| 303. |
Inwhich direction will the sun appear to set if the observe is inside the water of a pond? Refractive index of water , mu = 1.33. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :For a setting sun the incident rays graze along the surface of water i.e. <a href="https://interviewquestions.tuteehub.com/tag/angle-875388" style="font-weight:bold;" target="_blank" title="Click to know more about ANGLE">ANGLE</a> of incidence = `90^(@)`.<br/> `therefore "According in Snell.s law",` <br/> `mu_(w) = (sini)/(sinr) or, 1.33 = (sin90^(@))/(sinr)` <br/> `or, "" sinr = (1)/(1.33) or, 1.33 = sin48.75^(@)` <br/> `therefore "" r = 48.75^(@)` <br/> Therefore, to see the setting sun the <a href="https://interviewquestions.tuteehub.com/tag/observe-582964" style="font-weight:bold;" target="_blank" title="Click to know more about OBSERVE">OBSERVE</a> in water should <a href="https://interviewquestions.tuteehub.com/tag/look-544452" style="font-weight:bold;" target="_blank" title="Click to know more about LOOK">LOOK</a> at an angle of `48.75^(@)` with the normal.</body></html> | |
| 304. |
A pendulum clock is set to give correct time at the sea level. The clock is moved to a hill station at an altitude 'h' above sea level. In order to keep correct time on the hill station which one of the following adjustments is required? |
| Answer» <html><body><p>The length of the <a href="https://interviewquestions.tuteehub.com/tag/pendulum-1149901" style="font-weight:bold;" target="_blank" title="Click to know more about PENDULUM">PENDULUM</a> has to be reduced <br/>The length of the pendulum has to be <a href="https://interviewquestions.tuteehub.com/tag/increased-1040402" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASED">INCREASED</a> <br/>The mass of the pendulum has to be increased <br/>The mass of the pendulum has to be increased </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :a</body></html> | |
| 305. |
What was the book about? |
| Answer» <html><body><p>televisions<br/>geography<br/>school<br/>It was her <a href="https://interviewquestions.tuteehub.com/tag/grandfather-1010896" style="font-weight:bold;" target="_blank" title="Click to know more about GRANDFATHER">GRANDFATHER</a>'s diary</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 306. |
The power factor varies between |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a> to 1<br/>1 to 2<br/>2 to 2.5<br/>10 to 100</p>Answer :A</body></html> | |
| 307. |
"CC"l_(4) is inert towards hydrolysis but SiCl_(4) is readily hydrolysed because : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/carbon-16249" style="font-weight:bold;" target="_blank" title="Click to know more about CARBON">CARBON</a> cannot <a href="https://interviewquestions.tuteehub.com/tag/expand-454996" style="font-weight:bold;" target="_blank" title="Click to know more about EXPAND">EXPAND</a> its octat but silicon can expand its octet<br/>Ionisation potential of carbon is higher than silicon<br/>Carbon <a href="https://interviewquestions.tuteehub.com/tag/forms-11384" style="font-weight:bold;" target="_blank" title="Click to know more about FORMS">FORMS</a> <a href="https://interviewquestions.tuteehub.com/tag/double-441870" style="font-weight:bold;" target="_blank" title="Click to know more about DOUBLE">DOUBLE</a> and <a href="https://interviewquestions.tuteehub.com/tag/triple-712945" style="font-weight:bold;" target="_blank" title="Click to know more about TRIPLE">TRIPLE</a> bonds<br/>Electronegativity of carbon is higher than that of silicon</p>Solution :Carbon can not increase valency but Si can increase its covalency due to presence of vaccant d-orbital.</body></html> | |
| 308. |
निम्नलिखित में से कौन-सा प्रकथन सत्य है?1.0.25m फोकस दूरी के उत्तल लेंस की क्षमता 4 डाइऑप्टर होती है 2.0.25 m फोकस दूरी के उत्तल लेंस की क्षमता -4 डाइऑप्टर होती है3.0.25 m फोकस दूरी के अवतल लेंस की क्षमता 4 डाइऑप्टर होती है 4.0.25 m फोकस दूरी के अवतल लेंस की क्षमता - 4 डाइऑप्टर होती है |
| Answer» <html><body><p>1और 2<br/>2 और 3<br/>3 और 4<br/>1 और 4</p>Answer :D</body></html> | |
| 309. |
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emfof 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cmlength of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k Omegais put in series with it, which is shorted close to the balance point. The standard cell is then replaced bya cell of unknown emfe and the balance point found similarly, turns out to be at 82.3 cm length of the wire. (a) What is the value epsilon ? (b)What purpose does the high resistance of600 K Omega have ? (c) Is the balance point affected by this high resistance ? (d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0 V ? (e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermocouple) ? If not, how will you modify the circuit ? (f) Can we use above circuit to measure very small emf of the order of mV. (For example, emf obtained in thermocouple) ? If not, then what change would you make ? |
| Answer» <html><body><p></p>Solution :(a) We have , <br/> `(epsilon_(1))/(epsilon_(2)) = (l_(1))/(l_(2))` <br/> `therefore (1.02)/(epsilon) = (67.3)/(82.3)` <br/> `therefore epsilon = 1.02 xx (82.3)/(67.3)` <br/> `therefore epsilon = 1.247` V<br/> (b) Purpose of connecting high value resistance in <a href="https://interviewquestions.tuteehub.com/tag/secondary-638692" style="font-weight:bold;" target="_blank" title="Click to know more about SECONDARY">SECONDARY</a> circuit is to reduce the current through the galvanometer (and hence to save the galvanometer from the possible damage) when movable contact (jockey) is far from the balancing point. <br/> (<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>) No, because by connecting resistance in secondary circuit of a potentiometer,potential gradient existing on potentiometer wire does not changeand so balancing <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a> also does not change. <br/> (d) Yes, because`epsilon_(1) = kl_(1) = ((epsilon rho )/(R + L rho + r) ) l_(1) =` <br/>constant . <br/>`rArr`As r change , `l_(1) ` has to change. <br/> `rArr` Balancing poin will be affected. <br/> (e) No, because in order to obtain balancing point on the potentiometer wire, emf of driver cell (connected in primary circuit) must be greater than emf of unknown cell (connected in secondary circuit).<br/> Note : See the explanation given below. <br/> For primary circuit of a potentimeter, <br/> `I = (epsilon)/(R + L rho + r) "" `... (1) <br/> (Where `rho`= resistance per unit length of apotentiometer wire ) <br/>According to formula of potential gradient, <br/> k = `(epsilon rho)/(R + L rho + r)= ((epsilon)/(R + L rho + r) ) rho = I rho `.... (2) <br/> If emf `epsilon_(1) ` is balaned by length `l_(1)` of potentiometer wire then, <br/> `epsilon_(1) = kl_(1)` <br/> `therefore (epsilon_(1))_(max) = K(l_(1))_(max)= ` kL = L `rho L "" ` .... (3) <br/>From equation (1),<br/> `epsilon = IR+ IL rho + Ir `<br/>` therefore epsilon = IR+(epsilon_(1))_(max)+ Ir `<br/> ` therefore epsilon -(epsilon_(1))_(max) = IR+ Ir ` <br/> `therefore epsilon - (epsilon_(1))_(max) gt `0 <br/> `therefore epsilon gt (epsilon_(1))_(max) ` <br/> above is the required result . <br/> (f) In order to measure very small emf accurately, we <a href="https://interviewquestions.tuteehub.com/tag/require-11720" style="font-weight:bold;" target="_blank" title="Click to know more about REQUIRE">REQUIRE</a> to have very small potential gradient on the potentiometer wire. Value of k can be reduced considerably by connecting proper series resistance with potentiometer wire. Also by <a href="https://interviewquestions.tuteehub.com/tag/increasing-1040633" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASING">INCREASING</a> length of potentiometer wire, we can decrease the value of potential gradient.</body></html> | |
| 310. |
Capacitor is a device to store ......................... . |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/charge-914384" style="font-weight:bold;" target="_blank" title="Click to know more about CHARGE">CHARGE</a></body></html> | |
| 311. |
The mixture a pure liquid and a solution in long vertical column (i.e, horizontal dimensions lt lt vertical dimensions ) Produce diffusion of solute particles and hence refractive index gradient along the vertical dimesion . A ray of light entering the column I at right angles to the vertical is deviated from its original path. Find the deviated from its original path. Find the deviation in travelling a horizontal distance d lt lt h, the height of the coloumn. |
| Answer» <html><body><p></p>Solution :`rArr` As shown in the figure, consider an extremely narrow region of width dx, between the layers at distances x and x + dx , inside the extremely high cylindrical columnof liquid . <br/> `rArr` In above region, point B is on the level `bar(PQ)` , at height y from horizontal reference level, where <a href="https://interviewquestions.tuteehub.com/tag/refractive-2247079" style="font-weight:bold;" target="_blank" title="Click to know more about REFRACTIVE">REFRACTIVE</a> index is `mu` and gradient of refractiveindex. is `(dmu)/(dy)`. At this point, light ray `vec (AB)` is incident at angle `(180^@ - theta)` <br/> `therefore vec (AB)` , makes angle `(180^@ - theta)`with normal`M_1N_1` drawn on horizontal surface `bar(PQ)` , which becomes angle of incidence. ) If there would not be any gradient of refractive index, ray `vec(AB)` would have come to point B.. But here refractive index increases with the decreases in height and so at level `bar(RS)` height is (y - dy) andrefractive indes is `(mu + dmu)` which is more than `mu`. Hence, ray `vec(AB)`bends towards normal `M_1 N_1` and so it advances from B to C. Thus, ray `vec(BC)` becomes refracted light ray whichmakes angle `{180^@ - (theta + d theta)}` with normal `M_1 N_1`. Now applying Snell.s law atpoint B ion above figure, we get, <br/> `mu <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> sin (180^@ - theta) = (mu + d mu) xx sin { 180^@ - {theta + d theta)}` <br/>`because mu sin theta = (mu + d mu) sin (theta+ d theta)` <br/> `thereforemu sin theta = (mu + d mu )(sin theta cos d theta + cos theta sin d theta)` <br/> ` therefore mu sin theta = mu sin theta cos (d theta) + mu cos theta sin (d theta) + d(mu ) sin theta cos (d theta) + (d mu) cos theta sin (d theta)` <br/> `rArr` Since `d theta` is extremely small, we can take sin `(d theta)` <br/> `= d ( theta) cos (d theta) =<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>` <br/> `mu sin theta = (mu sin theta xx 1) + ( mu cos theta xx d theta)` <br/> `therefore0 = mu cos theta d theta + (d mu) sin theta ( because d theta cos theta d mu ` are negligible) <br/> `therefore ( d mu) sin theta = - mu cos theta d theta` <br/> `therefore (d mu ) sin theta = - mu cos theta d theta` <br/> `therefore ( dmu ) tan theta = -mu d theta`.........(1) <br/> `rArr` For the right angled `<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a>` BEB. in the figure<br/> `tan(180^@-B) = (BE)/(EB.)` <br/> `therefore tan theta= (dx)/(dy)`..........(2) <br/> `rArr` From equation (1) and (2,) <br/> `(d mu)(dx/dy) = - mu d theta` <br/> `therefore d theta = - (1)/(mu) (d mu)/(dy) dx` <br/> `rArr` Taking <a href="https://interviewquestions.tuteehub.com/tag/integration-11527" style="font-weight:bold;" target="_blank" title="Click to know more about INTEGRATION">INTEGRATION</a> on both the sides, <br/> `int_(0)^(@) = - (1)/(mu) ((dmu)/(dy)) int_(0)^(d) dx` <br/> `(therefore`During horizontal displacement `mu " and " (dmu)/(dy)` remain constant) <br/> `therefore theta = - (1)/(mu) ((dmu)/(dy))` d<br/> `rArr` Above equation gives required result .</body></html> | |
| 312. |
There are n cells in the circuit of figure. The emf and internal resistance of each cell are E and r respectively. The points A and B in the circuit divide the circuit into n and (N-n)cells. The current in the circuit is: |
| Answer» <html><body><p>`epsilon/r`<br/>`epsilon/r`<br/>`epsilon/r`<br/>Zero</p>Answer :A</body></html> | |
| 313. |
A copper wire of length 0.5 m and cross sectional area 10mm^(2) , carries a current perpendicular to a magnetic field of strength 10gauss. As a result, the wire experiences a force of 0.06 N. What is the drift velocity of free electrons in the copper wire? Take the number density of electrons= 8 xx 10^(28) m^(-3) and atomie weight of copper= 63.5 . |
| Answer» <html><body><p></p>Solution : Length of copper <a href="https://interviewquestions.tuteehub.com/tag/wire-1457703" style="font-weight:bold;" target="_blank" title="Click to know more about WIRE">WIRE</a>, `l = 0.5` m <br/>Area of cross section, `A = 10 mm^(2) = 10 xx 10^(-6) m^(2)` <br/>Number density of electrons, `n = 8 xx 10^(28) m^(-3)` <br/>Total charge inside the wire is given by <br/> `<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a> = <a href="https://interviewquestions.tuteehub.com/tag/ne-1112263" style="font-weight:bold;" target="_blank" title="Click to know more about NE">NE</a> xx ` Volume of wire <br/> `= ne xx Al` <br/> ` = 8 xx 10^(28) xx 1.6 xx 10^(-19) xx 10 xx 10^(-6) xx 0.5` <br/> ` = 6.4 xx 10^(4) C` <br/> Let `v_(d)` be the drift <a href="https://interviewquestions.tuteehub.com/tag/speed-1221896" style="font-weight:bold;" target="_blank" title="Click to know more about SPEED">SPEED</a> of electrons. Then, magnetic force on the wire due to external magnetic field B will be <br/> ` F = q v _(D) B sin theta` <br/> Here, `theta = 90^(@) , B = 10 " gauss " = 10 xx 10^(-4) T = 10^(-3) T`<br/> ` rArr "" F = qv_(D) B sin theta` <br/> ` rArr "" 0.06 = 6.4 xx 10^(4) xx v_(D) xx 10^(-3) xx 1` <br/> ` rArr "" v_(D) = (0.06)/(6.4 xx 10^(4) xx 10^(-3)) = 9.4 xx 10^(-4) ` m/s</body></html> | |
| 314. |
The point charges q_(A)=3 muC and q_(B)=-3 muC are located 20 cm apart in vaccum. What is the electric field at the midpoint O of the line AB joining the two charges? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :(a) `5.4 xx10^(6) <a href="https://interviewquestions.tuteehub.com/tag/nc-576183" style="font-weight:bold;" target="_blank" title="Click to know more about NC">NC</a>^(-1)` <a href="https://interviewquestions.tuteehub.com/tag/along-1974109" style="font-weight:bold;" target="_blank" title="Click to know more about ALONG">ALONG</a> OB <br/> (b) `8.1 xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>` alongOA</body></html> | |
| 315. |
An aeroplane is flying at a velocity of "900 km h"^(-1) loops a vertical circular loop. If the maximum force pressing the pilot against the seat is five times his weight, what would be the diameter (in m) of the loop? [g=10ms^(-2)] |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/3125-307035" style="font-weight:bold;" target="_blank" title="Click to know more about 3125">3125</a></body></html> | |
| 316. |
How many different combinations of three equal resistors can be made ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/three-708969" style="font-weight:bold;" target="_blank" title="Click to know more about THREE">THREE</a> <br/>four<br/>five<br/>six</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 317. |
In a Young.s double slit experiment , a monochromatic source if wavelength lambda is used to illmunate the two slits S_(1) and S_(2) The slitsS_(1)and S_(2)are identicaland sourceS is placedsymmetricallyas shown .Interference pattern is observed on a screen at a distanceD fromthe centreof slit . The distancebetween the slitsis d . If theresultantintensityat Pis same as that O , then the distance OP cannot be |
| Answer» <html><body><p>`(lambdaD)/d` <br/>`(2lambdaD)/d`<br/>`(3lambdaD)/d`<br/>`(1.5lambdaD)/d`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 318. |
In a Young.s double slit experiment , a monochromatic source if wavelength lambda is used to illmunate the two slits S_(1) and S_(2) The slitsS_(1)and S_(2)are identicaland sourceS is placedsymmetricallyas shown .Interference pattern is observed on a screen at a distanceD fromthe centreof slit . The distancebetween the slitsis d .If the source is movedby up by a very small disatncey_(0) , the centralmaximawill shift |
| Answer» <html><body><p>Upby `(y_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)d)/(x_(0))`<br/>Down by `(y_(0)d)/(x_(0))`<br/>Up by `(y_(0)D)/(x_(0))`<br/>Down by `(y_(0)D)/(x_(0))`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 319. |
In an a.c. circuit the e.m.f. € and the current (i) al any instant are-given respectively by e = E_(0)sin omegat I = l_(0)sin (omegat-phi) The average power in the circuit over one cycle of a.c. is |
| Answer» <html><body><p>`E_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)l_(0)`<br/>`(E_(0)l_(0))/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`<br/>`(E_(0)l_(0))/(2)sinphi`<br/>`(E_(0)l_(0))/(2)cosphi`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 320. |
In a Young.s double slit experiment , a monochromatic source if wavelength lambda is used to illmunate the two slits S_(1) and S_(2) The slitsS_(1)and S_(2)are identicaland sourceS is placedsymmetricallyas shown .Interference pattern is observed on a screen at a distanceD fromthe centreof slit . The distancebetween the slitsis d .If the sizeof slit S_(1)isslightly decreased , them |
| Answer» <html><body><p>Intensity at <a href="https://interviewquestions.tuteehub.com/tag/central-407816" style="font-weight:bold;" target="_blank" title="Click to know more about CENTRAL">CENTRAL</a> maximawill ramain same <br/>Intensity at <a href="https://interviewquestions.tuteehub.com/tag/sentral-2266181" style="font-weight:bold;" target="_blank" title="Click to know more about SENTRAL">SENTRAL</a> maximawill increase <br/>Intensity at <a href="https://interviewquestions.tuteehub.com/tag/first-461760" style="font-weight:bold;" target="_blank" title="Click to know more about FIRST">FIRST</a> minima will slightly increase from <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a> . <br/>Intensity at first minima will remain zero . </p>Answer :C</body></html> | |
| 321. |
...... changes in polarization |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a> <br/>Intensity<br/>Wavelength <br/><a href="https://interviewquestions.tuteehub.com/tag/phase-22748" style="font-weight:bold;" target="_blank" title="Click to know more about PHASE">PHASE</a> </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 322. |
A horizontal wire of length 0.2m long carries a current of 3A. Find the magnitude of the magnetic field, which can support the weight of the wire. Mass per unit length of wire is 2 xx 10^(-3) kg m^(-1) |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`6.53 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)T`</body></html> | |
| 323. |
Pendulum after some time becomes slow in motion and finally slopesdue to: |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/air-852233" style="font-weight:bold;" target="_blank" title="Click to know more about AIR">AIR</a> <a href="https://interviewquestions.tuteehub.com/tag/friction-21545" style="font-weight:bold;" target="_blank" title="Click to know more about FRICTION">FRICTION</a> <br/>mass of pendulum <br/>earth's gravity<br/>none of these.</p>Solution :Due to air friction the <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> of pendulum keeps on decreasing and hence its amplitudealso keeps on decreasing. <br/> Correctchoice is (a).</body></html> | |
| 324. |
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omegamaintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for every moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 Omegais put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsi and the balance point found similarly, turns out to be at 82.3 cm length of the wire. Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :No. If `<a href="https://interviewquestions.tuteehub.com/tag/epsi-2615801" style="font-weight:bold;" target="_blank" title="Click to know more about EPSI">EPSI</a>` is <a href="https://interviewquestions.tuteehub.com/tag/greater-476627" style="font-weight:bold;" target="_blank" title="Click to know more about GREATER">GREATER</a> than the <a href="https://interviewquestions.tuteehub.com/tag/emf-970036" style="font-weight:bold;" target="_blank" title="Click to know more about EMF">EMF</a> of the driver cell of the potentiometer, there will be no <a href="https://interviewquestions.tuteehub.com/tag/balance-891682" style="font-weight:bold;" target="_blank" title="Click to know more about BALANCE">BALANCE</a> point on wire AB.</body></html> | |
| 325. |
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omegamaintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for every moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 Omegais put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsi and the balance point found similarly, turns out to be at 82.3 cm length of the wire. Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The circuit, as it is would be unsuitable, because the <a href="https://interviewquestions.tuteehub.com/tag/balance-891682" style="font-weight:bold;" target="_blank" title="Click to know more about BALANCE">BALANCE</a> <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a> (for `epsi` of the order of a few mV) will be very close to the end A and the percentage error in measurement will be very large. The circuit is modified by putting a suitable resistor R in <a href="https://interviewquestions.tuteehub.com/tag/series-1201802" style="font-weight:bold;" target="_blank" title="Click to know more about SERIES">SERIES</a> with the wire AB so that potential drop across AB is only slightly greater than the emf to be measured. Then, the balance point will be at <a href="https://interviewquestions.tuteehub.com/tag/larger-1067345" style="font-weight:bold;" target="_blank" title="Click to know more about LARGER">LARGER</a> length of the wire and the percentage error will be much smaller</body></html> | |
| 326. |
What is dispersion of light? What is its cause? |
| Answer» <html><body><p></p>Solution :`N_P=<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>` Transformer ration K = 100<br/> (a) Number of <a href="https://interviewquestions.tuteehub.com/tag/turns-1429088" style="font-weight:bold;" target="_blank" title="Click to know more about TURNS">TURNS</a> in secondary <a href="https://interviewquestions.tuteehub.com/tag/coil-921250" style="font-weight:bold;" target="_blank" title="Click to know more about COIL">COIL</a> <br/> `N_S=N_PxxK=100xx100 = 10000`<br/> (b) Input power `(P_p)` = Input voltage `(V_P)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>` <a href="https://interviewquestions.tuteehub.com/tag/current-940804" style="font-weight:bold;" target="_blank" title="Click to know more about CURRENT">CURRENT</a> in primary `(I_P)` <br/> `1100 = 220xxI_PimpliesI_P=5A`<br/> (c) `V_S/V_P=N_S/N_P = ("Output Voltage")/("Input Voltage")`<br/> `implies V_S=5/I_SxxV_P=KxxV_P` <br/> `=220xx100 = 22000 V`<br/><br/> (d) `("Input Current")/("Output Current ")=I_P/I_S=N_S/N_P` <br/>` implies 5/I_S = 100 " or " I_S=0.05A`<br/> (e) For an ideal transformer, <br/> Power in secondary = Power in primary <br/> `impliesPs = 1100W`</body></html> | |
| 327. |
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omegamaintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for every moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 Omegais put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsi and the balance point found similarly, turns out to be at 82.3 cm length of the wire. What is the value epsi ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/epsi-2615801" style="font-weight:bold;" target="_blank" title="Click to know more about EPSI">EPSI</a> = 1.25 <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>`</body></html> | |
| 328. |
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omegamaintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for every moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 Omegais put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsi and the balance point found similarly, turns out to be at 82.3 cm length of the wire. What purpose does the high resistance of 600 K Omega have ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :to <a href="https://interviewquestions.tuteehub.com/tag/reduce-1181332" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCE">REDUCE</a> <a href="https://interviewquestions.tuteehub.com/tag/current-940804" style="font-weight:bold;" target="_blank" title="Click to know more about CURRENT">CURRENT</a> through the galvanometer when the <a href="https://interviewquestions.tuteehub.com/tag/movable-1104539" style="font-weight:bold;" target="_blank" title="Click to know more about MOVABLE">MOVABLE</a> contact is <a href="https://interviewquestions.tuteehub.com/tag/far-459491" style="font-weight:bold;" target="_blank" title="Click to know more about FAR">FAR</a> from the balance point.</body></html> | |
| 329. |
Define the term self inductance andwrite its S.I. unit. Obtain the expressioin for the mutual inductance of two long co-axial solenoids S_(1) and S_(2) wound one over the other, each of length L and radii n_(1) and r_(2) and n_(1) and n_(2) number of turns per unit lenght, when a current I is set up in the outer solenoids S_(2). |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Definitaioin of <a href="https://interviewquestions.tuteehub.com/tag/self-1199734" style="font-weight:bold;" target="_blank" title="Click to know more about SELF">SELF</a> inductance and its SI unit. <br/> Derivation of expression for mutual inductance. <br/> Self inductance of a coil equals,the magnitude of the magnetic flux, linked with it, when a unit current flows through it. <br/> Alternaively <br/> Self inductance, of a coil, equals the magnitude of the emf induced in it, when the current in the coil, is changing at a unit rate. <br/> SI unit: henry/ (webre/ampere)/(ohm second)<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DBT_SM_PHY_XII_EX_17_E01_027_S01.png" width="80%"/> <br/> When current `I_(2)` is passed through coil `S_(2)` it in <a href="https://interviewquestions.tuteehub.com/tag/turns-1429088" style="font-weight:bold;" target="_blank" title="Click to know more about TURNS">TURNS</a> set up a magnetic flux through <br/> `S_(1)=bar(omega)_(1)=(n_(1)l)(theta r_(1)^(2))(B_(2))` <br/> `bar(omega)_(1)=(n_(1)I)(pir_(1)^(2))(mu_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)n_(2)I_(2))` <br/> `bar(omega)_(1)=mu_(0)n_(1)n_(2)I_(2)pir_(1)^(2)lI_(2)` <br/> But `bar(omega)_(1)=M_(12)I_(2)` <br/> `impliesM_(12)=mu_(0)n_(1)n_(2)pir_(1)^(2)l`</body></html> | |
| 330. |
The speed v reached by a car of mass m in travelling a distance x, driven with constant power P, is given by |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>=(3xP)/(m)`<br/>`v=((3xP)/(m))^(1//<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`<br/>`v=((3xP)/(m))^(1//3)`<br/>`v=((3xP)/(m))^(2)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 331. |
A dice is placed with its one edge parallel to the principal axis between the principal focus and the centre to the curvature of a concave mirror. Then the image has the shape of |
| Answer» <html><body><p>Cube<br/>Cuboid <br/><a href="https://interviewquestions.tuteehub.com/tag/rectangular-1180623" style="font-weight:bold;" target="_blank" title="Click to know more about RECTANGULAR">RECTANGULAR</a> parallelopiped<br/>Spherical</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 332. |
A uniform electricfieldalongthex - axisis givenas,vecE =(200 hati)N* C^ (- 1 ) ,forxgt0=(-200 hati )N *C^ (- 1 ) ,forxlt0Acylinderof length20cmandradius5 cmhas itscentreattheorigin and axisalongthex - axisis placedin vacuum.findout(i)the electricfluxacrosseachof itscircularfaces, (ii)the fluxacrossitscurvedsurface,(iii)the fluxacrossitscentreoutersurfaceand (iv)thenet chargeenclosedby it. |
| Answer» <html><body><p></p>Solution :Radiusof thecylinder, `r =5 cm=0.05 m` <br/>`therefore`Areaofeachcircularface,<br/>` S =pi r ^ 2=3.14xx (0.05 ) ^ 2m ^ 2 ` <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XII_P1_U01_C02_SLV_060_S01.png" width="80%"/><br/>Asthelengthofthecylinderis20 cmor0.2 m, thetwocircularfacesareat`<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>=+ 0.1m andx=- 0.1m`.<br/>Theareavectorsrepresentingthe right and the leftcircularfacesare ` vec S _ <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>=hati(pi r ^ 2)andvec S _ 2=-hati(pi r ^ 2 )`respectively.<br/> theelectricfieldsat <a href="https://interviewquestions.tuteehub.com/tag/thepositionsof-3202244" style="font-weight:bold;" target="_blank" title="Click to know more about THEPOSITIONSOF">THEPOSITIONSOF</a> thesecircular facesare,` vec E _ 1=(200 hati)N * <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> ^(- 1andvec E _ 2=(-200 hati)N * C ^(- 1 ) `respectively.<br/> (i)The electricfluxacross therightcircularface, <br/>` phi _1=vec E _ 1 *vec S _ 1=(200 hati)*hati(pi r ^ 2 )` <br/>`=200 xx3.14 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(0.05 ) ^ 2=1.57N*m^ 2*C ^( -1 )`<br/> Similarly,for theleftcircularface,<br/>` phi _ 2=vec E _2 *vecS _ 2 =(- 200hati)*(-hatipir ^ 2 ) ` <br/>`= 200xx 3.14xx (0.05) ^ 2=1.57N *m ^ 2C ^ (- 1 ) ` <br/> `thereforephi _ 1= phi _ 2` i.e.,equalfluxpassesacrosseachof thetwofaces.<br/>(ii) The curvedsurface iseverywhereparallel to theelectric fieldvector.<br/>`therefore` Theelectricfluxlinkedwiththecurved surface=0 ` .<br/> (iii)Theentireoutersurfaceconsistsof thetwocircularfacesand thecurvedsurface. so, thefluxlinkedwithentiresurface,<br/> `thereforephi=phi _ 1+phi_ 2+0=1.57 + 1.57+ 0`<br/>` = 3.14 N*m^ 2*C ^(- 1 ) `<br/>(iv)From Gauss' theorem , netflux`phi=( q ) /(in _0) ` . So theenclosedchargeis,<br/>`q=phiin_ 0=3.14xx8.854 xx 10 ^(- 12 )` <br/>`=2.78 xx 10 ^(-11 ) `C</body></html> | |
| 333. |
A particlemoves in the planeaccording to thelaw x = kt, y = kt (1-alphat) , whenk and alphaare positive constants, and t is the time Find: (a) the equationof the particle's trajectoryy(x). (b) the velocityv andtheacceleration a of thethe point asa funtionof time . |
| Answer» <html><body><p></p>Solution :(a) `y = <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> - <a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a> (x^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(k^(2))` , (b) `bar(a) = (dbar(v))/(dt) = <a href="https://interviewquestions.tuteehub.com/tag/2alpha-1836743" style="font-weight:bold;" target="_blank" title="Click to know more about 2ALPHA">2ALPHA</a> k hat(j)`</body></html> | |
| 334. |
Assertion : Heart can be assumed as electric dipole. Reason : Its ELOF are just same like a normal dipole. |
| Answer» <html><body><p>If both <a href="https://interviewquestions.tuteehub.com/tag/assertion-384238" style="font-weight:bold;" target="_blank" title="Click to know more about ASSERTION">ASSERTION</a> and <a href="https://interviewquestions.tuteehub.com/tag/reason-620214" style="font-weight:bold;" target="_blank" title="Click to know more about REASON">REASON</a> are <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> and reason is the <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a> of assertion.<br/>If both assertion and reason are true but reason is not the correct explanation of assertion.<br/>If assertion is true but reason is false.<br/>If both assertion and reason are false.</p>Answer :A</body></html> | |
| 335. |
Find the power of a biconvex lens (mu = 1.5) with air on the left side and water (mu = 1.33) on the right side. The power of lens in air is+ 10 D : |
| Answer» <html><body><p>3.2 D<br/>4.4 D<br/>6.67 D<br/>2.4 D</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 336. |
In the following circuit (figure6.22) find the potential difference across the capacitors 1, 2, and 3. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`V_(1)=6V,V_(2)=<a href="https://interviewquestions.tuteehub.com/tag/2v-300496" style="font-weight:bold;" target="_blank" title="Click to know more about 2V">2V</a>,V_(3)=20V`</body></html> | |
| 337. |
The energy stored in a capacitor of capacitance Chaving a charge Q under a potential V is |
| Answer» <html><body><p>a. `1/2 Q^(2)<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>`<br/><a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>. `1/2C^(2)V`<br/><a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>. `1/2 Q^(2)/V`<br/>d. `1/2QV`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 338. |
What is self inductance of a coil ? Write its SI unit . Obtain the expression for energy store a in an inductor. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/self-1199734" style="font-weight:bold;" target="_blank" title="Click to know more about SELF">SELF</a> inductance of a coil is numerically equal to the emf induced in the coil due to unit rate of change of current in it. The SI unit of self inductance in henry (H). <br/> <a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> dw be the work <a href="https://interviewquestions.tuteehub.com/tag/done-2591742" style="font-weight:bold;" target="_blank" title="Click to know more about DONE">DONE</a> to establish a current I in the coil in a time dt. <br/> Then `(dw)/(dt)=|<a href="https://interviewquestions.tuteehub.com/tag/epsilon-973455" style="font-weight:bold;" target="_blank" title="Click to know more about EPSILON">EPSILON</a>|I "" epsilon` is the induced emf <br/> since `epsilon=-L (dI)/(dt)` <br/> `(dw)/(dt)=L(dI)/(dt)I` <br/> `dw=L(dI)/(canceldt)canceldtI` <br/> dw=LI dI <br/> The <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> work done in establishing the current is , <br/> `W=int dw=int_0^I L I dI =L[I^2/2]_0^1` <br/> `W=1/2LI^2` <br/> Thus the magnetic P.E. stored in an inductor of self inductance L carrying a current I is <br/> `W=1/2LI^2`</body></html> | |
| 339. |
Assume that the runner in sample question 3 completes the race in 1 inure and 20 seconds. Find her average speed and the magnitude of her average velocity. |
| Answer» <html><body><p></p>Solution :Average speed is <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> divided by leaped time Since the length of the track is 500 m, the runner's average speed was `(500 m)//(<a href="https://interviewquestions.tuteehub.com/tag/80-337972" style="font-weight:bold;" target="_blank" title="Click to know more about 80">80</a> s) =6.3 m//s`. However, since her <a href="https://interviewquestions.tuteehub.com/tag/displacement-956081" style="font-weight:bold;" target="_blank" title="Click to know more about DISPLACEMENT">DISPLACEMENT</a> was zero, her average velocity was zero <a href="https://interviewquestions.tuteehub.com/tag/also-373387" style="font-weight:bold;" target="_blank" title="Click to know more about ALSO">ALSO</a> : `v= Delta s//Delta t = (0 m)//(80s)= 0 m//s`.</body></html> | |
| 340. |
The aperture of an electron microscope is 0.02, the accelerating potential is 10^(4) V Find the dimensions of details which may be resolved with the aid of this instrument. |
| Answer» <html><body><p><br/></p>Solution :We <a href="https://interviewquestions.tuteehub.com/tag/use-1441041" style="font-weight:bold;" target="_blank" title="Click to know more about USE">USE</a> the expression for the resolving power of a microscope (66.8), putting `sinu=0.02`. We can <a href="https://interviewquestions.tuteehub.com/tag/find-11616" style="font-weight:bold;" target="_blank" title="Click to know more about FIND">FIND</a> the wavelength from the nonrelativistic formula, since the kinetic <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> of the electron of <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> keV is much <a href="https://interviewquestions.tuteehub.com/tag/less-1071906" style="font-weight:bold;" target="_blank" title="Click to know more about LESS">LESS</a> than its rest energy which is 510 keV.</body></html> | |
| 341. |
How does electroscope work ? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/use-1441041" style="font-weight:bold;" target="_blank" title="Click to know more about USE">USE</a> the white <a href="https://interviewquestions.tuteehub.com/tag/paper-1146067" style="font-weight:bold;" target="_blank" title="Click to know more about PAPER">PAPER</a> strips and fold the strips into half so that you make a mark of fold. Open the strip and iron it lightly as shown in <a href="https://interviewquestions.tuteehub.com/tag/figure-987693" style="font-weight:bold;" target="_blank" title="Click to know more about FIGURE">FIGURE</a>. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KPK_AIO_PHY_XII_P1_C01_E01_009_S01.png" width="80%"/><br/> Hold the strip by pinching it at the fold. You would notice that the <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> halves <a href="https://interviewquestions.tuteehub.com/tag/move-548879" style="font-weight:bold;" target="_blank" title="Click to know more about MOVE">MOVE</a> apart. This shows that the strip has acquired charge on ironing. <br/> When you fold it into half, both the halves have the same charge. Hence they repel each other.</body></html> | |
| 342. |
Two particles having charges Q_(1) and Q_(2) , when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled. Find the force between the particles. |
| Answer» <html><body><p></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> = (1)/(4 <a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a> epsilon_(0)) (Q_(1)Q_(2))/(<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>^(2))` <br/> If the distance is <a href="https://interviewquestions.tuteehub.com/tag/educed-7679860" style="font-weight:bold;" target="_blank" title="Click to know more about EDUCED">EDUCED</a> byhalf and two particles of charges are doubled. <br/> F. = `(1)/(4pi epsilon_(0)) (2Q_(1)2Q_(2))/((r//2)^(2)) = (1)/(4pi epsilon_(0)) (4Q_(1)Q_(2))/((r^(2)//4)^(2))` <br/> `F. = (1)/(4pi epsilon_(0)) (16(Q_(1)Q_(2)))/(r^(2)) = 16 [ (1)/(4pi epsilon_(0)) (Q_(1)Q_(2))/(r^(2))]` <br/> F. = 16 F</body></html> | |
| 343. |
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth's magnetic Meld at the place is 0.36 G and the angle of dip 18 zero. What is the total magnetic feld on the normal bisector of the magnet at the same distance as the mull-point (i.e., 14 cm) from the centre of the magnet? At rull points, field due to a magnet 1s equal and opposite to the horizontal component of earth's magnetic field.) |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/null-17167" style="font-weight:bold;" target="_blank" title="Click to know more about NULL">NULL</a> point of a given bar magnet is that point in the horizontal surface at which magnetic field due to that magnet is nullified by horizontal component of Earth.s magnetic field. <br/> For null points `P_1 and P_2` of a <a href="https://interviewquestions.tuteehub.com/tag/short-642706" style="font-weight:bold;" target="_blank" title="Click to know more about SHORT">SHORT</a> bar magnet on its axis. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KPK_AIO_PHY_XII_P1_C05_E02_041_S01.png" width="80%"/> <br/> We know that magnetic field of a bar magnet on its axis is in the direction of its dipole moment (in above <a href="https://interviewquestions.tuteehub.com/tag/figure-987693" style="font-weight:bold;" target="_blank" title="Click to know more about FIGURE">FIGURE</a>, towards right). <br/> Now, as shown in the diagram, if `P_1 and P_2` are the null points on the axis of given bar magnet then at these points, <br/> `B_(a) = B_(<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>)` <br/> `therefore (mu_(0))/( 4 pi) ((2m) /(r^3) )= 0.36b G ""...(1)` <br/> For the points `P_3 and P_4` of a short bar magnet on its equator located at distance r from its centre. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KPK_AIO_PHY_XII_P1_C05_E02_041_S02.png" width="80%"/> <br/> We know that magnetic field of a bar magnet is in the direction, opposite to its dipole moment (in above figure, towards left). <br/> At points `P_3 and P_4` (and also at all the points on the circle of radius r in the equatorial plane), magnetic field due to bar magnet is, <br/> `B_(e)= ((mu_0 )/(r^3) )` <br/> `= (1)/(2) ((mu_(0) )/( 4 pi )) ((2m)/( r^3) )` <br/> `= (1)/(2) B_a` <br/> `= (1)/(2) (0.36 )` [From equation (1)] <br/> `therefore B_(e) = 0.18 G` <br/> Now, if <a href="https://interviewquestions.tuteehub.com/tag/resultant-1187362" style="font-weight:bold;" target="_blank" title="Click to know more about RESULTANT">RESULTANT</a> magnetic field at points `P_(3) and P_(4)` (and also at all the points on a circle of radius r in the equatorial plane) is `overset(to) (B_R)` then, <br/> `overset(to) (B_R) = overset(to) (B_(e) ) + overset(to)(B_h) ` <br/> `therefore B_R = B_e + B_h ""(because "Here", overset(to) (B_e) || overset(to) (B_h) )` <br/> `= 0.18 + 0.36` <br/> `therefore B_R = 0.54` G <br/> Note : At null points of a bar magnet, resultant magnetic field is zero only along horizontal direction because vertical component of magnetic field of Earth is not balanced at the null point. This is because bar magnet is kept on the horizontal surface.</body></html> | |
| 344. |
Digital circuits can be made by repetitive use of |
| Answer» <html><body><p>OR <a href="https://interviewquestions.tuteehub.com/tag/gates-1004041" style="font-weight:bold;" target="_blank" title="Click to know more about GATES">GATES</a><br/>NOT gates <br/>AND gates<br/>NAND gates</p>Answer :D</body></html> | |
| 345. |
A nucleus of mass number A has a mass defect Deltam. Give the formula, for the binding energy per nucleon of this nucleus. |
| Answer» <html><body><p></p>Solution :Binding <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> <a href="https://interviewquestions.tuteehub.com/tag/nucleon-1126010" style="font-weight:bold;" target="_blank" title="Click to know more about NUCLEON">NUCLEON</a> `E_(bn)=(Deltam.c^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/A`</body></html> | |
| 346. |
A nonconducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its centre - (a) increases as r increases, for r lt R (b) decreases as r increases, for 0 lt r lt oo (c ) decreases as r increases, for R lt r lt oo (d) is dicontinuous at r = R |
| Answer» <html><body><p>a, c<br/>c, d<br/>a, b<br/>b, d</p>Answer :A</body></html> | |
| 347. |
A 500 g teapot and an insulated thermos are in a 20^(@)C room. The teapot is filled with 1000 g of the boiling water. 12 tea bags are then placed into the teapot. The brewed tea is allowed to cool to 80^(@)C, then 250 g of the tea is poured from the teapot into the thermos. The teapot is then kept on an insulated warmer that transfers 500 cal//min to the tea. Assume that the specific heat of brewed tea is the same as that of pure water, and that the tea bags have a very small mass compared to that of the water, and a negligible effect on the temperature. The specific heat of teapot is 0.17 J//gK and that of water is 4.18 J//gK. The entire procedure is done under atmospheric pressure. There are 4.18 J in one calorie. After the tea is added to the thermos, the temperature of the liquid quickly falls from 80^(@)C to 75^(@)C as it reaches thermal equilibrium with the thermos flask. What is the heat capacity of the thermos? |
| Answer» <html><body><p>`9.5 J//K`<br/>`14 J//k`<br/>`95 J//K`<br/>`878 J//K`</p>Solution :(a) `250gmxx4.18xx(80-75)=(<a href="https://interviewquestions.tuteehub.com/tag/ms-549331" style="font-weight:bold;" target="_blank" title="Click to know more about MS">MS</a>)(75-20)` <br/> `:.(ms)=95 J//k` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>)Granite (maximum <a href="https://interviewquestions.tuteehub.com/tag/specific-1220917" style="font-weight:bold;" target="_blank" title="Click to know more about SPECIFIC">SPECIFIC</a> heat) <br/> (c) as Rate `=500cal//min.:.5min=2500`cal. <br/> `2500xx4.18=(750xx4.18+500xx0.17)Deltatheta` <br/> <a href="https://interviewquestions.tuteehub.com/tag/rightarrow-623803" style="font-weight:bold;" target="_blank" title="Click to know more about RIGHTARROW">RIGHTARROW</a> `Deltatheta=3.24` final temperature`=83.24^(@)`C</body></html> | |
| 348. |
A 500 g teapot and an insulated thermos are in a 20^(@)C room. The teapot is filled with 1000 g of the boiling water. 12 tea bags are then placed into the teapot. The brewed tea is allowed to cool to 80^(@)C, then 250 g of the tea is poured from the teapot into the thermos. The teapot is then kept on an insulated warmer that transfers 500 cal//min to the tea. Assume that the specific heat of brewed tea is the same as that of pure water, and that the tea bags have a very small mass compared to that of the water, and a negligible effect on the temperature. The specific heat of teapot is 0.17 J//g K and that of water is 4.18 J//g K. The entire procedure is done under atmospheric pressure. There are 4.18 J in one calorie. An alternative method for keeping the tea hot would be, to place the teapot on a 10 pound block that has been heated in an oven to 300^(@)C. A block of which of the following substances would best be able to keep the tea hot? |
| Answer» <html><body><p>copper (specific heat `=0.39 J//gK)`<br/><a href="https://interviewquestions.tuteehub.com/tag/granite-1010923" style="font-weight:bold;" target="_blank" title="Click to know more about GRANITE">GRANITE</a> (specific heat`=0.79 J//gK)`<br/>iron (specific heat `=0.45 J//g K)`<br/>pewter (specific heat `=0.17 J//g K)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :(a) `250gmxx4.18xx(80-75)=(ms)(75-20)` <br/> `:.(ms)=95 J//k` <br/> (b)Granite (maximum specific heat) <br/> (c) as Rate `=500cal//min.:.5min=2500`cal. <br/> `2500xx4.18=(750xx4.18+500xx0.17)<a href="https://interviewquestions.tuteehub.com/tag/deltatheta-2053554" style="font-weight:bold;" target="_blank" title="Click to know more about DELTATHETA">DELTATHETA</a>` <br/> <a href="https://interviewquestions.tuteehub.com/tag/rightarrow-623803" style="font-weight:bold;" target="_blank" title="Click to know more about RIGHTARROW">RIGHTARROW</a> `Deltatheta=3.24` final temperature`=83.24^(@)`C</body></html> | |
| 349. |
A 500 g teapot and an insulated thermos are in a 20^(@)C room. The teapot is filled with 1000 g of the boiling water. 12 tea bags are then placed into the teapot. The brewed tea is allowed to cool to 80^(@)C, then 250 g of the tea is poured from the teapot into the thermos. The teapot is then kept on an insulated warmer that transfers 500 cal//min to the tea. Assume that the specific heat of brewed tea is the same as that of pure water, and that the tea bags have a very small mass compared to that of the water, and a negligible effect on the temperature. The specific heat of teapot is 0.17 J//g K and that of water is 4.18 J//g K. The entire procedure is done under atmospheric pressure. There are 4.18 J in one calorie. If, after some of the tea has been transferred to the thermos (as described in the passage), the teapot with its contents (at a temperature of 80^(@) C)was placed on the insulated warmer, what would be the temperature at the end of this 5 minute period?(Assume that no significant heat transfer occurs with the surroundings) |
| Answer» <html><body><p>`80.7^(@)` C<br/>`82.5^(@)` C<br/>`83.2^(@)`C<br/>`<a href="https://interviewquestions.tuteehub.com/tag/95-342378" style="font-weight:bold;" target="_blank" title="Click to know more about 95">95</a>.2^(@)`C</p>Solution :(a) `250gmxx4.18xx(80-75)=(ms)(75-20)` <br/> `:.(ms)=95 J//k` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>)Granite (<a href="https://interviewquestions.tuteehub.com/tag/maximum-556915" style="font-weight:bold;" target="_blank" title="Click to know more about MAXIMUM">MAXIMUM</a> specific heat) <br/> (c) as Rate `=500cal//min.:.5min=2500`cal. <br/> `2500xx4.18=(750xx4.18+500xx0.17)Deltatheta` <br/> Rightarrow `Deltatheta=3.24` <a href="https://interviewquestions.tuteehub.com/tag/final-461168" style="font-weight:bold;" target="_blank" title="Click to know more about FINAL">FINAL</a> temperature`=83.24^(@)`C</body></html> | |
| 350. |
In a transistor, ihe emitter circuit resis- tance is 100Omega and the collector resistance is 100Omega. The power gain, if the emitter and collector currents are assumed to be equal, will be |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>></p>Solution :If `I_(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>)~~I_(E)rArr beta~~1` <br/> `therefore A_(P)=beta^(2)((R_(L))/(R_(i)))~~((R_(L))/(R_(i)))=(100xx10^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))/(100)=<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>` <br/></body></html> | |