InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 651. |
When an object is placed at a distance of 25 cm from a mirror, the magnification is `m_(1)`. The object is moved 15cm farther away with respect to the earlier position, and the magnification becomes `m_(2)`. If `m_(1)//m_(2)=4` , then calculate the focal length of the mirror. |
|
Answer» We know that `m=-(v)/(u)=(f)/(f-u)` Here, `m_(1)=(f)/(f-(-25))=(f)/(f+25)` and `m_(2)=(f)/(f-(-25-15))=(f)/(f+40)` Since ` (m_(1))/(m_(2))=4, ` therefore `(f+40)/(f+25)=4` `f+40=4f+100` or `f=-20 cm` The negative sign shows that the mirror is concave. |
|
| 652. |
Two thin lenses of focal lengths 20cm and 25cm are placed in contact. The effective power of the combination isA. `(1)/(9)` diopterB. 9 diopterC. 6 diopterD. 45 diopter |
| Answer» Correct Answer - B | |
| 653. |
The lateral magnification of the lens with an object located at two different position `u_(1)` and `u_(2)` are `m_(1)` and `m_(2)` , respectively. Then the focal length of the lens isA. `f=sqrt(m_(1)m_(2))(mu_(2)-mu_(1))`B. `f=sqrt(m_(1)m_(2))(u_(2)-u_(1))`C. `((u_(2)-u_(1)))/(sqrt(m_(1)m_(2)))`D. `((u_(2)-u_(1)))/((m_(2))^(-1)-(m_(1))^(-1))` |
|
Answer» Correct Answer - d. For lens `(1)/(f)=(1)/(v_(1))-(1)/(u_(1))` `(u_(1))/(f)=(u_(1))/(v_(1))-1rArrm_(1)=(v_(1))/(u_(1))=(f)/(u_(1)+f)` And `m_(2)=(f)/(u_(2)+f)` `(1)/(m_(2))-(1)/(m_(1))=((u_(2)-mu_(1)))/(f) rArr f=((mu_(2)-mu_(1)))/((m_(2))^(-1)-(m_(1))^(-1))` |
|
| 654. |
Find the lateral magnification produced by the combination of lenses shown in Figure. |
|
Answer» If lenses are placed in contact, the focal length of single equivalent length can be given as `(1)/(f)=(1)/(f_(1))+(1)/(f_(2))=(1)/(10)-(1)/(20)=(1)/(20)` `rArr f=+20` Now using lens formula considering system of lenses as a single lens, we get `(1)/(v)-(1)/(u)=(1)/(f)` `:. (1)/(v)=(1)/(-10)=(1)/(20)` `(1)/(v)=(1)/(20)-(1)/(10)=(-1)/(20)rArrv=-20cm` Now using magnification formula, we get `m=(-20)/(-10)=2` |
|
| 655. |
A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principal axes coinciding. A beam of light travelling parallel to the principal axis and having a beam diameter 5.0 mm, is incident on the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of the emergent beam. |
|
Answer» Correct Answer - 2.5 mm |
|
| 656. |
At `27^@C` two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate (i) the final temperature of the gas, (ii) change in its internal energy, and (iii) the work done by the gas during this process. |
|
Answer» Correct Answer - (i) `189K (ii) -2767J (iii) 2767J` (i) For adiabatic process `T_(1)V_(1)^(gamma -1) = T_(2)V_(2)^(gamma-1)` `implies 300V^(5/3-1) = T_(2).(2V)^(5//3-1) implies T_(2) = 189K` (ii) Change in internal energy `DeltaU = nC_(V)DeltaT = n(3)/(2)R(T_(2)-T_(1)) = 2 xx (3)/(2)xx 8.314 xx (189 - 300) = -2767J` (iii) Work done by gas = `(P_(1)V_(1)-P_(2)V_(2))/(gamma-1) = (nR(T_(1) - T_(2)))/(gamma-1) = 2767J` |
|
| 657. |
In a simple microscope, if the final image is located at infinity then its magnifying power isA. `D/f`B. `1+D/f`C. `f/D`D. `Dxxf` |
|
Answer» Correct Answer - A `D/f` |
|
| 658. |
An object 15cm high is placed 10cm from the optical center of a thin lens. Its image is formed 25cm from the optical center on the same side of the lens as the objectA. 2.5cmB. 0.2cmC. 16.7cmD. 37.5cm |
|
Answer» Correct Answer - d. ` (I)/(O)=(v)/(u)` `(I)/(15)=(-25)/(-10)` `I=15xx2.5cm=37.5cm` |
|
| 659. |
An object of height 4 cm is kept to the left of and on the axis of a converging lens offocal length 10cm at a distance of 15cm from lens. A plane mirror is placed inclined at `45^(@)` to the lens axis, 10 cm to the right of the lens.Find the position and size of the image formed by the lens and mirror combination. Trace the path ofthe rays forming the image. |
|
Answer» Correct Answer - 20 cm from the mirror, 8 cm |
|
| 660. |
An extended object of size 2mm is placed on the principal axis of a converging lens of focal length 10cm. It is found that when the object is placed perpendicular to the principal axis the image formed is 4mm in size. The size of image when it is placed along the principal axis is `"____________"` mm. |
|
Answer» Correct Answer - 8 When object is placed perpendicular to the principal axis, image size `=mxx(` object size) or `4=mxx2rArrm=2` When object is placed along principal axis, image size`=m^(2)` (object size) `=4xx2=8mm` |
|
| 661. |
An extended object is placed at a distance of 5.0 cm from a convex lens of focal length 8.0 cm. (a) Draw the ray diagram (to the scale) to locate the image and from this, measure the distance of the image from the lens, (b) Find the position of the image from the lens formula and see how close the drawing is to the correct result. |
|
Answer» Correct Answer - A Draw yourself |
|
| 662. |
A concave convex figure lens made of glas `(mu=1.5)` has surface of radii 20 cm and 60 cm. a. Locate the image ofan object placed 80 cm to the left of the lens along the principal axis. B. A similar lens is placed coaxially at distanc of 160 cm right of it. Locate the position of the image. |
|
Answer» The focal length of the lens is given by `1/f=(mu-1)(1/R_1-1/R_2)` `=(1.5-1)(1/(20cm)-1/(60 cm))=1/(60cm)` or `f=60 cm a. for the image formed by lens `u=-80` cm so that `1/v=1/u+1/f` `=1/(-80cm)+1/(60cm)=1/(240cm)` or `v=240 cm` The first image `I_1` would form 240 cm to the right of the first lens. b. The second lens intercepts the converging beam as suggest by the ure. the image `I_1` acts as a virtual source of the second lens. For the image formed by this lens `u=240cm-160cm=+80` cm so that `1/v=1/u+1/f` ` =1/(80cm)+1/(60cm)=7/(240cm)` or `v=34.3cm` the fiN/Al image is formed 34.3 cm to the right of the second lens. |
|
| 663. |
In a given process on an ideal gas, `dW=0` and `dQ` is negative, then for the gas:A. the temperature will decreaseB. the volume will decreaseC. the pressure will remain constantD. the temperature will increase |
|
Answer» Correct Answer - A If `dW = 0 , dQ lt 0 "then " dU lt 0` `implies` The temperature will decrease |
|
| 664. |
A compound microscope has an objective of focal length `2.0 cm` and an eye-piece of focal length `6.25cm` and distance between the objective and eye-piece is `15cm`. If the final image is formed at the least distance vision `(25 cm)`, the distance of the object form the objective is |
|
Answer» Here, `f_(0) = 2.0 "cm", f_(e) = 6.25 "cm" , u_(0) = ?` (a) ` v_(e)=-25 "cm" because (1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e)) therefore (1)/(u_(e)) = (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25) - (1)/(6.25) = (-1-4)/(25) = (-5)/(25) implies u_(e) = -5"cm"` As distance between objective and eye piece = `15"cm" , v_(0) = 15-5 = 10"cm"` `because (1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0)) implies (1)/(u_(0)) = (1)/(v_(0)) - (1)/(f_(0)) = (1)/(8.75) - (1)/(20) = (2-8.75)/(17.5) implies u_(0) = (-17.5)/(6.75) = -2.59"cm"` Magnifying power = `(v_(0))/(|u_(0)|) xx [1 + (D)/(f_(e))] = (v_(0))/(|u_(0)|)xx (D)/(|u_(e)|) = (8.75)/(2.59)xx (25)/(6.25) = 13.51` |
|
| 665. |
A telescope of diameter `2m` uses light of wavelength `5000 Å` for viewing stars.The minimum angular separation between two stars whose is image just resolved by this telescope isA. `4xx10^(-4)rad`B. `0.25 xx10^(-6)rad`C. `0.31 xx10^(-6)rad`D. `5.0xx10^(-3)rad` |
|
Answer» Correct Answer - C Minimum angular separation `Deltatheta=(1)/(R.P.)=(1.22lambda)/(d)` `(1.22xx5000xx10^(-10))/(2)=0.3xx10^(-6)rad` |
|
| 666. |
In a vessel, as shown in Figure., point P is just visible when no liquid is filled in vessel through a telescope in the air. When liquid is filled in the vessel completely, point Q is visible without movint the vessel or telescope. Find the refractive index of the liquid. A. `(sqrt(14))/(3)`B. `(sqrt(85))/(5)`C. `sqrt(2)`D. `sqrt(3)` |
|
Answer» Correct Answer - b. Form similar triangles APB and CBD, point P will be just visible if `CD=4R` `musini=1xxsinr` `musin(angleCDQ)=1xxsin(angleCDB)` `muxx(R)/(Rsqrt(7))=1xx(2R)/(Rsqrt(20))rArr mu=(2sqrt(17))/(sqrt20)=(sqrt(85))/(5)` |
|
| 667. |
A convex lens forms an image of an object on a screen. The height of the image is 9 cm . The lens is now displaced until an image is again obtained on the screen. Then height of this image is 4 cm . The distance between the object and the screen is 90 cm.A. The distance between the two positions of the lens is 30 cmB. The distance of the object from the lens in its first position is 36 cmC. The height of the object is 6 cmD. The focal length of the lens is 21.6 cm |
|
Answer» Correct Answer - A,D Height of object = `sqrt(" H of image" xx H "of image "2) = sqrt(I_(1)I_(2)) = sqrt(9 xx4)` Height of object = 6 cm `m_(1) = (I_(1))/(O) = (v_(1))/(u) implies (9)/(6) = (v_(1))/(u_(1))` `u_(2) = v_(1) = (3)/(2) v_(1) therefore X = (3)/(2)u_(1) - u_(1) = (u_(1))/(2) = X` ` D = 90 = u_(2) + u_(1) = (3)/(2) u_(1) + u_(1)` `u_(1) = 36 implies X = (36)/(2) = 18` `therefore` Distance between two position of lens = X = 18 Focal length `f = (D^(2) - V^(2))/(4D) = (90^(2) - 18^(2))/(4 xx 90) = 21.6 "cm"` |
|
| 668. |
Statement-`1`: Temperature of a rod is increased and again cooled to same initial temperature then itsfinal length is equal to original length provided there is no deformation take place. Statement-`2`: For a small temperature change , length of a rod varies as `l = l_(0)(1+alphaDeltaT)` provided `alpha DeltaT ltlt1`. Here symbol have their usual meaning.A. Statement -`1` is True , Statement -`2` is True , Statement -`2` is correct explanation for Statement - `1`B. Statement -`1` is True , Statement -`2` is True , Statement -`2` is not correct explanation for Statement - `1`C. Statement -`1` is True , Statement -`2` is False.D. Statement -`1` is False , Statement -`2` is True. |
| Answer» Correct Answer - B | |
| 669. |
Statement-`1`: Coolant coils are fitted at the top of a refrigerator , for formation of convection current. Statement-`2`: Air becomes denser on cooling.A. Statement -`1` is True , Statement -`2` is True , Statement -`2` is correct explanation for Statement - `1`B. Statement -`1` is True , Statement -`2` is True , Statement -`2` is not correct explanation for Statement - `1`C. Statement -`1` is True , Statement -`2` is False.D. Statement -`1` is False , Statement -`2` is True. |
| Answer» Correct Answer - A | |
| 670. |
In a home experiment , Ram brings a new electric kettle with unknown power rating . He puts `1` litre water in the kettle and switches on . But to his dismay , the temperature becomes constants at `60^(@)C` after some time. The room temperature is `20^(@)C`. Ram gets bored and switches off the kittle . He sees that during first `20`s water cools down by `2^(@)C`. Which is the best graph for temperature `v//s` time?A. B. C. D. |
| Answer» Correct Answer - B | |
| 671. |
In a home experiment , Ram brings a new electric kettle with unknown power rating . He puts `1` litre water in the kettle and switches on . But to his dismay , the temperature becomes constants at `60^(@)C` after some time. The room temperature is `20^(@)C`. Ram gets bored and switches off the kittle . He sees that during first `20`s water cools down by `2^(@)C`. What is the wattage of the kettleA. `840W`B. `1W`C. `100W`D. `420W` |
|
Answer» Correct Answer - B Ans (D) `(DeltaQ)/(t) =(1 xx 4200 xx 2)/(20)J//sec = 420W` |
|
| 672. |
Consider a hypothetical situation where we are comparing the properties of two crystals made of atom `A` and atom `B`. Potential energy `(U) v//s` interatomic separation (r) graph for atom `A` and atom `B` is shown in figure (i) and (ii) and respectively. It is seen that the potential energy can reach a maximum value of `U_(T)` at temperature `T= 10K`. if `r_(1)` and `r_(2)` are `0.9999 r_(0) "and" 1.0003 r_(0)` for atoms of crystal `A`, its approximate coefficient of linear expansion can be :-A. `4xx10^(-5)//K`B. `1 xx 10^(-5)//K`C. `2 xx 10^(-5)//K`D. `3 xx 10^(-5) //K` |
|
Answer» Correct Answer - C We have `Deltar_("avg") = ((1.0003-0.9999)/(2)) r_(0)` (from equilibrium position) `therefore (Deltar)/(r_(0)) = alpha DeltaT implies alpha = 2 xx 10^(-5)//K` |
|
| 673. |
Consider a hypothetical situation where we are comparing the properties of two crystals made of atom `A` and atom `B`. Potential energy `(U) v//s` interatomic separation (r) graph for atom `A` and atom `B` is shown in figure (i) and (ii) and respectively. When we heat the crystal of either atoms ,the atom undergo oscillation. Choose correct statement for atoms of crystal `A`A. Their equilibrium position remains unchanged but average separation decreasesB. Their equilibrium position remains unchanged but average separation increasesC. Their separation at equilibrium position as well as average separation increasesD. Their separation at equilibrium position decreases but average separation increases |
|
Answer» Correct Answer - B The equilibrium remain unchanged but average distance increases. |
|
| 674. |
The weight of a person is `60` kg . If he gets `10` calories of heat through food and the efficiency of his body is `28%`, then upto what height he can climb? Take g = `10m s^(-2)`A. `100` cmB. `1.96` cmC. `400` cmD. `1000` cm |
|
Answer» Correct Answer - B Amount of energy utilised in climbing mgh = `0.28 xx10xx4.2` ` implies h = (0.28xx10xx4.2)/(60xx10) = 1.96 xx 10^(-2)m = 1.96`cm |
|
| 675. |
250 g of water and equal volume of alcohol of mass 200 g are replaced successively in the same calorimeter and cool from `60^@C` to `55^@C` in 130 s and 67 s, respectively. If the water equivalent of the calorimeter is 10 g, then the specific heat of alcohol in `cal//g^@C` isA. `1.30`B. `0.67`C. `0.62`D. `0.985` |
|
Answer» Correct Answer - C Rate of cooling of water = Rate of cooling of alcohol `implies ((250+10) xx 1xx(5))/(130) = ((200s + 10)xx5)/(67)` `implies ` Specific heat of alcohol s = `0.62` |
|
| 676. |
A cube of side `2m` is placed in front of a concave mirrorr of focal length `1m` with its face `A` at a distance of `3m` and face `B` at a distance of `5m` form the mirrorr, The distance between the images of face `A` and `B` and height of images of `A` and `B` are respectively. A. `1m, 0.5,0.25m`B. `0.5m,1m,0.25m`C. `0.5m,0.25m,1n`D. `0.25m,1m,0.5m` |
|
Answer» Correct Answer - D For A `u=-3m` `v_(1)=?,` `f=-1m` `(1)/(u_(1))=(1)/(f)-(1)/(u)=(1)/(-1)-(1)/(-3)=(1)/(3)=1=-(2)/(3)` or `v_(1)=-(3)/(2)` For B `(1)/(v_(2))=(1)/(-1)-(1)/(-5)` or `(1)/(v_(2))=(1)/(5)-1=-(4)/(5)` or `v_(2)=-(5)/(4)m` Now, `v_(1)-v_(2)=(3)/(2)-(-(5)/(4))` `=-(3)/(2)+(5)/(4)=-(1)/(4)m=-0.25m` Again, `(I_(1))/(Q)=-(v_(1))/(u)` or `I_(1)=-(v_(1))/(u)Q-((-3)/(2))((-1)/(3))=-1m` Again `(I_(2))/(Q)=-(v_(2))/(u)` or `I_(1)=(-(5)/(4))((1)/(-5))2= -0.5m` |
|
| 677. |
A conveying lens of focal length 15 cm and a converging mirror of focal length 20 cm are placed with their principal axes coinciding. A point source S is placed on the principal axis at a distance of 12 cm from the lens as shown in figure. It is found that the final beam comes out parallel to the principal axis. Find the separation between the mirror and the lens. |
|
Answer» Let us first locate the image of S formed byi the lens L. Here `u=-12cm and f=15 cm`. We have `1/v-1/u=1/f` `or 1/v=1/f+1/u` `=1/(15cm)-1/(12cm)` `or v=-60cm` The negative sign shows that the image is formed to the left of the lens as suggeted in the ure. The image `I_1` acts as the source for the mirror. The mirror forms and image `I_2` of the source `I_1`. This image I_2 then acts as the source for the lens and the fiN/Al beam comes out parallel to the principal axis. Clearly `I_2` must be at the focus of the lens. We have `I_1I_2=I_1L+L_2=60cm15cm=75cm`. Suppose the distance of the mirror from `I_2` is x cm. For the reflection from the mirror. `u=MI_1(=-(75+x)cm, v=-xcm =75cm` Suppose the distance of the mirror from `I_2` is cm. For the reflection from the mirror. `u=MI_1=-(75+x)cm,v=-xcm and f=-20cm` Using `1/v+1/u=1/f` `1/x+1/(75+x)=1/20` `or (75+2x)/((75+x)x)=1/20` `or x^2+35-1500=0` or `x=(-35+-sqrt(35xx35+4x1500))/2` This gives x=25 or -60. As the negatives sign has no physical meaning only positive sigh should be taken. Taking `x=25` the separatioin between the lens and the mirror is `(15+25)cm=40cm.` |
|
| 678. |
Calculate the focal length of the thin lens shown infigure. The pionts `C_1 and C_2` denote the centres of curvature. |
|
Answer» As is clear from the ure boththe radii of curvature are positive. Thus, `R_1=+10cm` and `R_2=+20cm.` The focal length is given by `1/f=(mu-1)(1/R_1-1/R_2)` `=(1.5-1)(1/(10cm)-1/(20cm))` `=0.5xx1/(20cm)=1/(40cm)` `or f=40cm.` |
|
| 679. |
An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens ofA. Small focal length and small diameterB. Small focal length and large diameterC. Large focal length and large diameterD. Large focal length and small diameter |
|
Answer» Correct Answer - C For astronomical refracting telescope Angular magnification is more for large focal length of objective lens `(M.P.=(f_(o))/(f_(e)))` Resolving power `=(d)/(1.22lambda)` Resolving power is high for large diameter . |
|
| 680. |
A prism has refracting angle of `60°` and its material has refractive index 1.5 and 1.6 for red and violet light respectively. A parallel beam of white light is incident on one face of the prism such that the red light undergoes minimum deviation. Find the angle of incidence (i) and the angular width `(theta)` of the spectrum obtained. Given: `sin (49^(@)) = 0.75^(@), sin(28^(@)) = 0.47, sin(32^(@)) = 0.53, sin(58^(@)) = 0.85` |
|
Answer» Correct Answer - `i=49^(@) ; theta = 9^(@)` |
|
| 681. |
Three closed vessels `A, B` and `C` are at the same temperature T and contain gases which obey the Maxwellian distribution of velocities. Vessel A contains only `O_(2), B` only `N_(2)` and `C` a mixture of equal quantities of `O_(2)` and `N_(2)`. If the average speed of the `O_(2)` molecules in vessel `A` is `V_(1)`, that of the `N_(2)` molecules in vessel `B` is `V_(2)`, the average speed of the `O_(2)` molecules in vessel `C` is (where `M` is the mass of an oxygen molecules)A. `(v_(1) + v_(2))//2`B. `v_(1)`C. `(v_(1)v_(2))^(1//2)`D. `sqrt(3kT//M)` |
| Answer» Correct Answer - B | |
| 682. |
During an experiment, an ideal gas is found to obey a condition `(p^2)/(rho) = "constant"`. (`rho` = density of the gas). The gas is initially at temperature (T), pressure (p) and density `rho`. The gas expands such that density changes to `rho//2`.A. The pressure of the gas changes to `sqrt2P`B. The temperature of the gas changes to `sqrt2T`.C. The graph of the above process on the `P-T` diagram is parabola.D. The graph of the above process on the `P-T` diagram is hyperbola. |
|
Answer» Correct Answer - B::D `(P^(2))/(rho)` = constant ` P = p(RT)/(M)` (Ideal gas equation) `implies (p_(2))/(rho) = (P)/(rho)((rhoRT)/(M)) = PT(R)/(M)` = constant `therefore` The graph of the above process on the `P-T` diagram is hyperbola. For the above process `((P_(2))/(rho))_(1) = ((P^(2))/(rho))_(2) implies (P_(2))/(rho) = (P_(2)^(2))/(rho//2) implies P_(2) = (P)/(sqrt(2)) .... (i)` and `P_(1)T_(1) = P_(2)T_(2) implies PT = (P)/(sqrt(2)) T_(2) implies T_(2) = sqrt(2) T .... (ii)` |
|
| 683. |
The emissive power of a black body at `T=300K` is `100Watt//m^(2)` consider a body B of area `A=10m^(2)` coefficient of reflectivity `r=0.3` and coefficient of transmission `t=0.5` its temperature is 300 K. then which of the followin is correct:A. The emissive power of `B` is `20W//m^(2)`B. The emissive power of `B` is `200W//m^(2)`C. The power emitted by `B` is `200"Watt"`D. The emissivity of `B` is =` 0.2` |
|
Answer» Correct Answer - B For body `AP = sigmaAT^(4)` `(P)/(A) = 100 = sigma xx 300^(4)` For body `B` `(P)/(A) = (sigmaT^(4)) in = (1-0.5-0.3) xx [sigmaxx300^(4)]` ` = 0.2 xx 100 = 20W//m^(2)` |
|
| 684. |
Certain perfect gas is found to obey `PV^(n)` = constant during adiabatic process. The volume expansion coefficient at temperature `T` isA. `(1-n)/(T)`B. `(1)/(1-n)T`C. `(n)/(T)`D. `(1)/(nT)` |
|
Answer» Correct Answer - B `PV^(n)` = constant & =`muRT implies V prop T^(((1)/(1-n)) implies(DeltaV)/(V) = ((1)/(1-n))(Delta T)/(T)` `implies` volume expansion coefficient = ` (DeltaV)/(VDeltaT) = (1)/((1-n)T)` |
|
| 685. |
The figure shows two thin rods , one made of aluminium `[alpha = 23 xx 10^(-6) (C^(@))^(-1)]` and the other of steel `[alpha = 12 xx 10^(-6) (C^(@))^(-1)]`. Each rod has the same length and the same initial temperature . They are attached at one end to two separate immovable walls. Temperature of both the rods is increased by the same amount , until the gap between the rods vanishes . Where do the rods meet the gap vanishes? A. The rods meet exactly at the midpoint.B. The rods meet to the right of the midpoint.C. The rods meet to the left of the midpoint .D. Information insufficient |
|
Answer» Correct Answer - B As `alpha_(At) gt alpha_("steel")` so expansion in aluminum rod is greater . |
|
| 686. |
Three identical rods of length 1 m each, having cross-sectional area of `1cm^2` each and made of aluminium, copper and steel, respectively, are maintained at temperatures of `12^@C`,`4^@C` and `50^@C`, respectively, at their separate ends. Find the teperature of their common junction. `[K_(Cu)=400 W//m-K,K_(Al)=200 W//m-K,K_("steel")=50 W//m-K]` |
|
Answer» ` R_(Al) = (L)/(KA)=(1)/(200xx10^(-4)) = 10^(4)/200` Similarly `R_(steel) = 10^(4)/50` and ` R_(copper)= (10^(4))/(400)` Let temperature of common junction = `T` then from Kirchoff,s current laws, `i_(Al) + i_(steel) +i_(Cu) =0` ` implies (T-12)/(R_(Al)) + (T-51)/R_(steel) + (T-4)/R_(Cu)=0` ` implies (T-12)200+(T-50)50 +(T-4)400=0` `implies 4(T-12)+ (T-50) + 8(T-4)=0` ` implies 13T=48+52+32=130` ` implies T=10^(@)C` |
|
| 687. |
`5`n , n and `5`n moles of a monoatomic , diatomic and non-linear polyatomic gases (which do not react chemically with each other ) are mixed at room temperature . The equivalent degree of freedom for the mixture is :-A. `(25)/(7)`B. `(48)/(11)`C. `(52)/(11)`D. `(50)/(11)` |
|
Answer» Correct Answer - D `f_(eq) = (f_(1)n_(1) + f_(2)n_(2) + f_(3)n_(3))/(n_(1) + n_(2)+ n_(3)) = ((5n) (3)(n)(5)+(5n)(6))/(5n + n+ 5n) = (50)/(11)` |
|
| 688. |
A particle moves in a circular path of radius 5 cm in a plane perpendicular to the principla axis of a xonvex mirror with radius of curvature 20cm. The object is 15cm in front of the mirror. Calculate the radius of the circular path of the image. |
|
Answer» `u=-15cm, f=(+20)/(2)=10cm` `(1)/(-15)+(1)/(v)=(1)/(10)` `(1)/(v)=(1)/(10)+(1)/(15)rArrv=6cm` `m=-(v)/(u)=(h_(2))/(h_(1))` `(-6)/(-15)=(h_(2))/(5)rArr h_(2)=2cm` So, radius of circulart path of the image is 2 cm. |
|
| 689. |
Radius of curvature of concave mirrorr is `40cm` and the size of image is twice as that of object, then the object distance isA. `60cm`B. `20cm`C. `40cm`D. `30cm` |
|
Answer» Correct Answer - D `f=(R)/(2)=20cm m=2` For real image`,m=-2` Now, `m=(f)/(f-u)implies-2=(-20)/(-20-u)impliesu=-30cm` For virtual image `,m=+2` So, `+2=(-20)/(20-u)impliesu=-10cm` |
|
| 690. |
A concave lens with unequal radii of curvature made of glass `(mu_(g)=1.5)` has a focal length of 40cm. If it is immersed in a liquid of refractive index `mu_(1)=2`, thenA. it behaves like a convex lens of 80cm focal lengthB. it behaves like a convex lens of 20cm focal lengthC. its focal length becomes 60cmD. nothing can be said |
|
Answer» Correct Answer - a. `-(1)/(40)=(1.5-1)((1)/(R_(1))-(1)/(R_(2)))` `(1)/(R_(1))-(1)/(R_(2))=-(1)/(20)` Now, `(1)/(f)=((1.5)/(2)-1)(-(1)/(20))` `=-(0.5)/(2)(-(1)/(20))` `=(1)/(80)` or `f=80cm` It behaves like a convex lens of focal length 80 cm. |
|
| 691. |
Let `barv,v_(rms) and v_p` respectively denote the mean speed. Root mean square speed, and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m. ThenA. No molecule can have speed greater than `v_("rms")`B. No molecule can have speed less than `(v_(p))/(sqrt(2))`C. `v_(p) lt overlinev lt v_("rms")`D. The average kinetic energy of a molecule is `(3)/(4) mv_(p)^(2)` |
|
Answer» Correct Answer - C::D `v_(p) = sqrt((2RT)/(M)) , overlinev = sqrt((8RT)/(piM)), v_(rms) = sqrt((3RT)/(M))` |
|
| 692. |
An opera glass `(`Galilean telescope ) measures `9cm` from the objective to the eyepiece. The focal length of the objective is `15 cm`. Its magnifying power isA. `2.5`B. `2//5`C. `5//3`D. `0.4` |
|
Answer» Correct Answer - A `f_(o)-f_(e)=9cm` and `f_(e)=f_(o)-9=15-9=6cm` `impliesm=(f_(e))/(f_(e))=(15)/(6)=2.5` |
|
| 693. |
The graph between sine of angle of refraction `(sin r)` in medium 2 and sine of angle of incidence `(sin i)` in medium indicates that `(tan 36^(@)~~(3)/(4))`A. Total internal reflection can take placeB. Total internal reflection cannot take placeC. Any of `(a)` and `(b)`D. Data is incomplete |
|
Answer» Correct Answer - B From graph, slop `= tan((2pi)/(10))=(sin r)/(sin i)` Also, `._(1)mu_(2)=(mu_(2))/(mu_(1))=(sini)/(sinr)=(1)/(tan((2ppi)/(10)))=(4)/(3)impliesmu_(2)gtmu_(1)` |
|
| 694. |
A ray of light is incident at an angle of `60^@` on the face of a prism having refracting angle `30^@.` The ray emerging out of the prism makes an angle `30^@` with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges.A. Normal to the face through which it emergesB. Inclined at `30^(@)` to the face through which it emergesC. Inclined at `60^(@)` to the face through which it emergesD. None of these |
|
Answer» Correct Answer - A We know that `delta=i+e-Aimplies e=delta+A-i` `=30^(@)+30^(@)-60^(@)=0^(@)` `:.` Emergent ray will be perpendicular to the face. Therefore if will make an angle of `90^(@)` with the face through which it emerges. |
|
| 695. |
In a certain spectrum produced by a glass prism od dispersive power `0.031`, it was found that `mu_(r ) = 1.645 and mu_(b) = 1.665`. What si the refractive index for yellow colour ? |
|
Answer» Here, `omega = 0.031 , mu_(r) = 1.645 mu_(v)= 1.665, mu_(y) = ?` `because omega=(mu_(v)-mu_(r))/(mu_(y)-1) = therefore mu_(y)-1 = (mu_(v) - mu_(r))/(omega) = (1.665-1.645)/(0.031) = (0.020)/(0.031) = 0.645 therefore mu_(y) = 0.645+1 = 1.645` |
|
| 696. |
A real object is placed in front of a convex mirror (fixed).The object is moving toward the mirror. If `v_(0)` is the speed of object and `v_(i)` is the speed of image, thenA. `v_(i) lt v_(0)` alwaysB. `v_(i) gt v_(0)` alwaysC. `v_(i) gt v_(0)` initially and then `v_(0) gt v_(i)`D. `v_(i) lt v_(0)` initially and then `v_(i) gt v_(0)` |
|
Answer» Correct Answer - A As object moves from infinity to the pole of convex mirror, image moves from focus to pole. So `v_(i) lt v_(0)` , always. |
|
| 697. |
For a concave mirror of focal length f, image is 2 times larger. Then the object distance from the mirror can beA. `(f)/(2)`B. `(3f)/(2)`C. `(f)/(4)`D. `(4f)/(3)` |
|
Answer» Correct Answer - A::B If image is virtual. Then, Suppose `u=-x`, then `v=+2x` From the mirror formula `(1)/(u)+(1)/(v)=(1)/(f)` or`(1)/(2x)-(1)/(x)=(1)/(-f)` or `x=(f)/(2)` If image is real, then suppose` u=-y, ` then `v=-2y` Again applying the mirror formula, `(1)/(-2y)-(1)/(y)=-(1)/(f)` or `(3)/(2y)=(1)/(f)rArr y=(3)/(2)f` |
|
| 698. |
If in the previous example, we use a diverging lens with a focal length `10.0cm` to form an image of the pencile kept 15 cm in front of the lens, locate and characterize the image. |
|
Answer» In accordance with cartesian sign convention, the given parameters are `f=-10cm, u=-15cm` From lens equation, we have `=(1)/(v)-(1)/(f)+(1)/(u)` ` =(1)/((-10))+(1)/((-15))=-(1)/(6),v=-6cm` Lateral magnification `m=(v)/(u)=(-6)/(-15)=0.40` The minus sign with image shows that the image is located on the side of the object. The magnification is positive and `m lt 1` , which shows that the image is upright and diminished. |
|