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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 38201. |
Voltage and current in an a.c. circuit are given by V=5 sin (100pit-pi/6) and I=4 sin (100pit+pi/6) then ………….. |
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Answer» voltage LEADS the CURRENT `30^@` `=pi/6-(-pi/6)` `=pi/3 = 60^@` |
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| 38202. |
Two inductors of inductance L_(1) and L_(2) are put in (a) series and (b) parallel with a large separation. Given L_(1)=6H and L_(2)=4H. Calculate the equivalent inductance. |
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Answer» <P> Solution :a. EQ. inductance `L_(s)=L_(1)+L_(2)=10H`b. Eq. inductance `L_(P)=(L_(1)L_(2))/(L_(1)+L_(2))=(6xx4)/(10)=2.4H` |
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| 38203. |
A charged bead is capable of sliding freely through a string held vertically in tension. An electric field is applied parallel to the string so that the bead stays at rest of the middle of the string. If the electric field is switched off momentarily and switched on |
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Answer» the bead moves downwards and stops as soon as the field is SWITCHED on |
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| 38204. |
Lencho was not happy to see the 70 pesos. |
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Answer» Correct |
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| 38205. |
The maximum speed that can be achieved without skidding by a car on a circular unbanked road of radius R and coefficient of static friction p is |
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Answer» `muRg` |
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| 38206. |
Select outputs Y of the combination of gates shown below for inputs A = 1 , B = 0 , A = 1 , B = 1 and A = 0 , B = 0 respectively. |
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Answer» `{:(( 0, 1, 1 )):}` |
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| 38207. |
The sequence of coloured bands in two carbon resistors R_1 and R_2 is (i) brown, green, blue and (ii) orange, black, green. Find the ratio of their resistances. |
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Answer» Solution :As per colour code FOLLOWED resistance of (i) RESISTOR `R_1` having brown, green and blue bands is `R_1 = 15 XX 10^6 Omega` (ii) Resistor `R_2` having orange, BLACK and green bands is `R2 = 30 xx 10^5 Omega` `R_1/R_2 = (15 xx 10^6)/(30 xx 10^5) = 5/1` |
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| 38208. |
U^(238) decays in 14 steps that together consitute the net reaction ._(92)U^(238)rarr._(82)^(206)Pb+_...+6vecv+8(._(2)^(4)He) (a) Fill in the blank in the above reaction. (b) The atomic masses of U^(238),Pb^(206) and He^(4) are 238.050783u. 205.97449u, and 4.002603u respectively. The energy equivalent of mass of one electron is 0.51 MeV. with this data calculate the Q value of the reaction. (c) U^(238) has a long half life of t=4.47xx10^(9) yr. the other members in the chain reaction have relatively small half lives. As soon as a nucleus of U^(238) decays the remaining steps are completed almost instataneously. with this assumption, calculate the power output from a 100kg block of pure U^(238). the density of U^(238) is 18700 Kg//m^(3). |
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Answer» (C) 0.01 WATT. |
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| 38209. |
Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements L,C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 11 for this frequency. |
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Answer» SOLUTION :As they are CORRECTED in the parallel combination `(1)/(Z)=(1)/(R )+((1)/(X_(L))-(1)/(X_(C )))` As the reactance `(X_(C )-X_(L))` is perpendicular to the ohmic resistance R, therefore we can write as `(1)/(Z)=sqrt((1)/(R^(2))+((1)/(X_(L))-(11)/(X_(C )))^(2))` `= sqrt((1)/(R^(2))+((1)/(omega L)-omega C)^(2))` At resonance, `omega = omega r = (1)/(sqrt(LC))` That means `(1)/(Z)` = minimum and thus Z = maximum. As Z is maximum, current will be minimum. current through INDUCTOR `I_("RMS")L=(V_("rms"))/(X_(L))=(230)/(50xx5)` `I_("rms")L=(230)/(250)=0.92 A` current through capacitor `I_("rms")C=(V_("rms"))/(X_(C ))=(V_("rms")xx omega C)/(1)` `= 230xx50xx80xx10^(-6)` `= 0.92A`. Current throught resistor `I_("rms")R=(I_("rms"))/(R )=(230)/(40)=5.75 A` |
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| 38210. |
An increase in intensity level of one decibel implies an increase in intensity of |
| Answer» ANSWER :C | |
| 38211. |
The work done in rotating a magnet from the direction of uniform field to the opposite direction of the field is W. Find work done in rotating the magnet from the field direction to half the maximum couple position |
| Answer» SOLUTION :`(W)/(4)(2-sqrt(3))` | |
| 38212. |
The sound level at a point P is 14 dB below the sound level at a point 1.0 m from a point source. The distance from the source to point P is |
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Answer» 4.0 m |
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| 38213. |
When a p-n junction diode is reverse biased, the flow of current across the junction is maninly due to …….. |
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Answer» diffusion of charges |
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| 38214. |
Two identical tuning forks vibrating at the same frequency 220 Hz are kept fixed at some distance apart. A listener runs parallel between the forks at a speed of 3 m/s so that he approaches one tuning fork and recedes from the other. Find the beat frequency observed by the listener (speed of the sound = 330 m/s) |
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| 38215. |
A person cannot see objects clearly 50 cm from him. Can you help him by telling the nature and focal length of lens which he must use to make his eye normal? |
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Answer» |
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| 38216. |
A long struight conductor carrying a current lies along the axis of a ring. The conductor will exert a force on the ring if |
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Answer» CARRIES a current |
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| 38217. |
In Fig. , a ball is launched with a velocity of magnitude 10.0 m/s, at an angle of 50.0^@to the horizontal. The launch point is at the base of a ramp of horizontal length d_1 = 6.00 m and height d_2 = 3.60 m. A plateau is located at the top of the ramp. (a) Does the ball land on the ramp or the plateau? When it lands, what are the (b) magnitude and (c) angle of its displacement from the launch point? |
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Answer» Solution :(a) Let m = `d_2//d_1 = 0.600` be the slope of the ramp ,so y = mx there , we choose our coordinate ORIGIN at the point of launch and use eq. thus ` y = tan (50.0^@) x- ((9.80 m//s^2)x^2)/(2(10.0 m//s)^2 (cos 50.0^@)^2) = 0.600 x ` which yield x = 4.99 m. this is less than `d_1` so the ball does land on the ramp (b) Using the value of x FOND in part (a) , we obtain`y = mx = 2.99 m `. thus , the pythagorean theorem yield a displacement MAGNITUDE of `sqrt( x^2 + y^2) = 5.82 m` (c ) The angle is , of course , the angle of the ramp : `tan^(-1) (m)` = 31.0^@`. |
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| 38218. |
The angular speed of the revolution of earth required so that the body on its surface at the equator would feel no weight is, (Radius of earth= 6400 km, g=9-8 m/s_2) |
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Answer» `1.723xx10^(-3)`rad/s |
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| 38219. |
(a) Fig. 9.04 shows a cross-section of a Tight pipe' made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.(b) What is the answer if there is no outer covering of the pipe ? |
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Answer» Solution : (a) Here refractive index of glass fibre `n_g = 1.68` and refractive index of OUTER covering `n_c = 1.44` Total internal reflection will take place inside the pipe if angle of incidence `i. gt i_( c)` , the critical angle, for glass fibre-outer covering interface. But `sin i_( c) =1/(n_(ge)) =n_( c)/n_(G) = 1.44/1.68 = 0.855 rArr i_( c) sin^(-1) (0.855) = 59^(@)` Hence, for total reflection `i^(.) gt 59^(@)`, or R, which has a value (90 -i.), should have a value `r le (90 -59)` or `r le 31^(@)`. Now according to Snell.s law `(sin i)/(sin r) = n_(g)` `therefore sin i = n_(g). sin r =(1.68) xx sin 31^(@) = 1.68 xx 0.5150 = 0.8652` `therefore i = sin^(-1) (8652) = 60^(@)` Hence, range of the ANGLES of incident rays with the axis of pipe = less than or equal to ± 60°. If there is no outer covering, then air will act as the covering and hence the critical angle `i_( c)^(.) = sin^(-1) (1/n_(g)) = sin^(-1) (1/1.68) = 36.5^(@)` Now, for rays incident along the axis of pipe i = 90° and applying Snell.s law `(sin i)/(sin r) = (sin 90^(@))/(sin r) = 1/(sin r) = n_(g) = 1.68`, we find that r= `36.5^(@)` Consequently, `i. = 90-r - 36.5 = 53.5^(@)`and it has a value greater than the critical angle i.c. Hence, all the rays will be totally reflected. |
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| 38220. |
Air that initially occupies 0.280 m at a gauge pressure of 103.0 kPa is expanded isothermally to a pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. It then returns to its initial pressure in a constant-volume process. Compute the network done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.) |
| Answer» SOLUTION :`1.12 XX 10^4J` | |
| 38221. |
Whatis thedistanceof closestapproachwhena 5MeVprotonapproachesa goldnucleus ? |
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Answer» Solution :USING` r_0=(1)/( 4 pi epsi_ 0) (Ze^2 )/(E ) `,we get `r_0 = ((9 xx 10^9 ) xx 79xx (1.6 xx 10^(-19) )^2)/( 5XX 10^6 xx 1.6 xx 10^(-19) )=228 xx 10^(-11) m` |
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| 38222. |
The magnetic field in plane e.m. wave is given by B_(y) = 2 xx 10^(7) sin (0.5 xx 10^(3)x+1.5 xx 10^(11)t) Tesla. The peak value of electric field is |
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Answer» `50 V m^(-1)` |
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| 38223. |
Given vec(A) = 2hat(i) + 4hat(j) -6hat(k). When a vector vec(B) is added to vec(A) , we get a unit vector along x-axis. Then vec(B) is : |
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Answer» `hat(i) + 2hat(j) - 3hat(K)` `vec(B)=-hati-4hatj+6hatk` |
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| 38224. |
Rainbows are classic example of the phenomenon of |
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Answer» Interference |
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| 38225. |
A ray of light is incident on a glass sphere of refractive index 3/2. What should be the angle of incidence so that the ray which enters the sphere does not come out of the spher? |
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Answer» `tan^(-1)(2//3)` |
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| 38226. |
Two wires having different densities are joined at x = 0. An incident wave y_1 = A sin(wt - k_1x) , travelling to the right in the wire is partly reflected and partly transmitted at x=0. If A_rand A_tbe the amplitudes of reflected and transmitted waves respectively, |
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Answer» `y_r = A_r sin (OMEGA t + k_1 X)` |
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| 38227. |
A plane electromagnetic wave travels in free space along X - direction. If the value of vec(B) (in tesla) at a particular point in space and time is 1.2xx10^(-8)hat(k). The value of vec(E ) (in V m^(-1)) at that point is |
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Answer» `1.2 HAT(J)` |
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| 38229. |
The apparent dip at a place 30^(@) away from the magnetic meridian is 60^(@) Calculate the true dip at the place |
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Answer» |
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| 38230. |
Calculate the work done when one mole of an ideal monoatomic gas is compressed adiabatically. The initial pressure and volume of the gas are 10^5N//m^2 and 6 litre respectively. The final volume of the gas is 2litres. Molar specific heat of the gas at constant volume is 3R/2. |
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Answer» Solution :For an adiabatic change `P_1V_1^(GAMMA)=P_2V_2^(gamma)` or `P_2=P_1(V_1/V_2)^(gamma)` Here `P_1=10^5 N//m^2, V_1` =6 litre , `V_2` =2 litreand for a monoatomicgas , we have `C_V=(3R)/2` and `C_P=(3R)/2+R=(5R)/2` or `gamma=C_P/C_V=(5R//2)/(3R//2)=5/3` Thus `P_2=10^5xx(6/2)^(5//3)=10^5xx(3)^(5//3)` We know that work done on gas `DeltaW`in adiabatic change is given by `DeltaW=(P_2V_2-P_1V_1)/(gamma-1)` `=(10^5xx(3)^(5//3)xx(2xx10^(-3))-10^5 xx(6xx10^(-3)))/((5//3)-1)` [Here `V_2`=2 litre `=2xx10^(-3)m^3`and `V_1`=6 litre= `6xx10^(-3)m^3` ] `=(2xx10^2[(3)^(5//3)-3])/(2//3)` `=(2xx10^2[6.19-3])/(2//3)=3xx10^2xx(3.19)` =957 J |
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| 38231. |
A tuning fork A is being tested using an accurate oscillator. It is found that they produce 2 beats per second when the oscillator reads 514 Hz and 6 beats per second when it reads 510 Hz. The actual frequency of the fork in Hz is |
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Answer» 512 Hz `f_(A)=(514+-2)Hz` `impliesf_(A)=516Hz` or 512 Hz for 2nd reading of oscillator `f_(A)=(510+-6)Hz` `impliesf_(A)=516Hz` or 504 Hz `implies` A has frequency of 516 Hz |
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| 38232. |
According to Rutherford's atomic model, the electron inside an atom are |
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Answer» stationary |
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| 38233. |
The electrostatic force on a small sphere of charge 0.4 muC due to another small sphere of charge -0.8 muC in air is 0.2 N. What is the force on the second sphere due to the first? |
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Answer» SOLUTION :(a) 12 cm (B) 0.2 N (ATTRACTIVE) |
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| 38234. |
When stopping potential is applied in an experiment of photoelectric effect, no photocurrent is observed. This means that |
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Answer» The EMISSION of photo ELECTRONS is stopped |
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| 38235. |
Charge q_2, is at the center of a circular path with radius r. work done in carrying charged q_1, once a round this equipotential path, would be. |
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Answer» `[1/(4 PI epsilon_0) * (q_1q_2) / R^2]` |
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| 38236. |
What was their first meal in two days? |
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Answer» MEAT loaves |
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| 38237. |
The device which easily closes or opens an electric circuit is called as__________. |
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Answer» JUNCTION diode |
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| 38238. |
A whistling engine is approaching a stationary observer with velocity 110 m/s. The velocity of sound is 330 m/s. The ratio of frequencies as heard by observer at the time of approaching and passing of engine is : |
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Answer» Solution :`v_(1) ((V)/(V - U_(s))) v and v_(2) = ((V)/(V + U_(s)))` V `therefore (v_(1))/(v_(2)) = (V + U_(s))/(V - U_(s)) = (330 + 110)/(330 - 110) = (v_(1))/(v_(2))= 2 ` HENCE the CORRECT CHOICE is (d). |
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| 38239. |
The time period of revolution of an electron in its ground state orbit in a hydrogen atom is 1.6xx10^(-16) s. The frequency of the electron in its first excited state (in s^(-1) ) is : |
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Answer» `7.8xx10^(14)` Suppose, time period is `T_(1)` in ground state and `T_(2)` in first excited state. `:.n_(1)=1` and `T_(1)=1.6xx10^(-16)s` and `n_(2)=2` and `T_(2)=?` `:(T_(2))/(T_(1))=((n_(2))/(n_(1)))^(3)""` (z=1 same ) `:.T_(2)=T_(1)XX((2)/(1))^(3)` `=1.6xx10^(16)xx8` `=12.8xx10^(-16)` `:.` FREQUENCY `v_(2)=(1)/(T_(2))=(1)/(12.8xx10^(-16))` `:.v_(2)=0.0781xx10^(16)` `:.v_(2)~~7.8xx10^(14)s^(-1)` |
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| 38240. |
Explain Polarisation by scattering. |
Answer» Solution :The light from a clear BLUE portion of the sky shows a rise and fall of intensity when VIEWED through a polaroid which is rotated. Because sunlight which has changed its direction on encountering the molecules of the earth.s atmosphere. This is called scattering. This is as shown in figurc (a).  he incident sunlight is UNPOLARISED. The dots stand for polarisation perpendicular to the plane of the figure. The double arrows show polarisation in the plane. Under the influence of the ELECTRIC field of the incident wave the electrons in the molecules acquire components of motion in both these directions. We have drawn an observer looking at `90^(@)` to the direction of the sun. Here, charges accelerating parallel to the double arrows do not radiate energy towards this observer since their acceleration has no transverse component. The radiation scattered by the molecule is therefore represented by dots. It is polarised perpendicular to the plane of the figure. This explains the polarisation of scattered light from the sky. In figure polarisation of the blue scattered light from the sky is shown. The scattering of light by molecules was investigated by C.V. Raman and his collaborators in the 1920s. Raman was awarded the Nobel prize for physics in 1930 for this WORK. |
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| 38241. |
Which of the following statements are true about acceleration due to gravity ? |
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Answer» G decreases in moving away from the centre if `r GT R` If `r lt R, g.="g"[(R-d)/(R )]` Where d is the depth below the surface of earth. If d=R, g=0 Further, due to ROTATION : `g.=g-Romega^(2) cos^(2)THETA` If `omega=0`, then g increases. |
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| 38242. |
The circumference of track is 256 m. If the maximum speed at which a car be driven safely along it is 25 m/s. Then angle of banking of the track is, g=9.8 m/s^2 |
| Answer» Answer :D | |
| 38243. |
A shell is fired from a canon with a velocity theta at angle with the horizontal direction. At the highest point in its path it explodes into two pieces of equal masses. One of the pieces retraces its path to the canon. The speed of the other piece immediately after the explosion is: |
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Answer» 3V COS `theta` Now `m_1=m_2=m/2` and `v_1=-v cos theta` `m/2(-v cos theta)+m/2v_2=mv cos theta` or`v_2=3v cos theta`. |
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| 38244. |
What is the force required to take away a flat circular plate of radius 2cm from the surface of water. (S.T. = 70 dynes/cm) |
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Answer» 400 dynes |
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| 38245. |
An ideal mono atomic gas is taken through the cyclic process shown in fig. Linear expansion from A to B following by adiabatic compression back to original state |
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Answer» Efficiency of cyclic process is zero `eta = ("work done")/("Heat given") = 1 - (Q_("released"))/(Q_("absorbed"))` Therefore there MUST be some heat absorbed in theprocess. During adiabatic process `DeltaQ = 0`, therefore during linear process heat will ENTRE and leaves SYSTEM at different times. |
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| 38246. |
A spherical conducting shell of inner radius r_1 and outer radiusr_2 ahs a charges 'Q' A charges 'q' is placed at the centre of the shell. (a)What is the surface charges density on the (i) inner surface (ii) outer surface of the shell? (ii)Write the expression for the electric field at a pointx gt r_2 from the centre of the shell. |
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Answer» |
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| 38247. |
The self-inductance of a coil having 200turns is 10 millihenry. Calculate the magnetic flux through the cross-section of the coil corresponding to current of 4 milliampere. Also determine the total flux linked with the coil. |
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Answer» SOLUTION :Total magnetic flux LINKED with the coil, `phi= LI= 10^(-2) H xx 4 xx 10^(-3)A` `phi= 4 xx 10^(-5)Wb` Magnetic flux through the cross-section `=(phi)/(200) = 2 xx 10^(-7)Wb` |
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| 38248. |
bomb moving with velocity 40i + 50j + 25k m/s explodes into pieces of mass ratio 1 : 4. If the small piece goes out with velocity 200i +70j -15k m/s, find the velocity of larger piece after explosion. |
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Answer» SOLUTION :Let 5m be mass of bomb. The largest PIECE will have mass 4m. Using principle of momentum CONSERVATION, `5m(40hati + 50hatj - 25hatk) =m(200hati + 70hatj-15hatk) + 4mvecv` Which gives `vecv =(180hatj + 140hatk)/4 =(45hatj + 35hatk)` |
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| 38249. |
Of metals and alloys, which have greater value of tempurature co.efficient of resistense? |
| Answer» Solution :METALS have greater VALUE of temperature co.efficient of resistance than alloys. | |
| 38250. |
Find the position and nature of the image of an object of height 3 cm when placed 60 cm from a mirror of focal length 15cm, when the mirror isconvex. |
| Answer» SOLUTION :VIRTUAL, erect of SIZE 0.6 cm, at a distance of 12 cm BEHIND the mirror | |