40048 + Interview Questions in Physics in Class 11

1.	Angular momentum is given by
Answer» `(I)/(OMEGA)` `TAU omega` `I omega` `( omegaI)/(2)` ANSWER :C

Discussion

2.	Relation between phase difference (Delta phi) and path difference (Deltax) is ......
Answer» `Delta phi= (2pi)/(LAMDA) Delta x` `Delta phi = 2pi lamda Delta x` `Delta phi = (2pi lamda)/(Delta x)` `Delta phi = (2 Deltax)/(lamda)` Solution :Wave-vector K indicates PHASE difference between two particles, separated by unit DISTANCE, along the directon of propagation of wave. Hence phase difference between two particles separated by distance `Delta x` WOULD be: `Delta phi = k (Delta x)` `therefore Delta phi = (2pi )/(lamda ) Delta x (because k = (2pi )/(lamda ))`

Discussion

3.	Two plano concave lenses of glass of refractive index 1.5. have radii of curvature 20cm and 30cm respectively. They are placed in contact with the curved surface towards each other and the space between them is filled with a liquid of refractive index 5//3. The focal length of the combination is
Answer» 6 `-92` 108 12 Answer :D

Discussion

4.	Product of 2.78 and 0.996 is divided by 1.527, the result has _________ significant figures
Answer» ONE two THREE four ANSWER :C

Discussion

5.	The angle between (vecA+vecB)and(vecA-vecB) can be
Answer» only `0^(@)` only `90^(@)` between `0^(@)` and `90^(@)` between `0^(@)` and `180^(@)` ANSWER :A::B::D

Discussion

6.	One end of a wire of radius r is sealedwith the end of another wire of radius 2r . This is used as a sonometer wire with tension T , keeping the junction at the mid-point of the vibrating length . If a stationary wave having a node at the junction is generated,find out the ratio between the number of loops in the two portions .
Answer» Solution :Let`rho_(1) and rho_(2) ` be the densities of thematerials of the TWO wires respectively . Mass PER unit length of the FIRST wire , ` m_(1) = pi r^(2) rho _(1)` Mass per unit length of the second wire , ` m_(2) = pi (2r)^(2) rho_(2) = 4 pi r^(2) rho _(2)` `:.(m_(1))/(m_(2)) = (rho_(1))/(4 rho_(2))` The junction is at the mid-point ofthe vibrating length 2L (say) . So , the vibrating length of each wire = l . The two wires arevibrating simultaneoulsy , so the frequency of vibration isthe same . Letit be n .Again, a NODE is generated at the junnction , so an integral number of loops is produced in each wire. if the number of loops in the two wires are p and q , respectively , then ` n = (p)/(2l) sqrt((T)/(m_(1))) and n = (q)/(2l) sqrt((T)/(m_(2)))` `:.(p)/(2l) sqrt((T)/(m_(1))) = (q)/(2l) sqrt((T)/(m_(2)))` or , ` (p)/(q) = sqrt((m_(1))/(m_(2))) = sqrt ((rho_(1))/(4 rho_(2))) = (1)/(2) sqrt((rho_(1))/(rho_(2)))` If the two wires are of the same material, then ` rho_(1) = rho_(2) ` . So , `(p)/(q) = (1)/(2)*`

Discussion

7.	The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it?
Answer» The forces may be acting radially from a point on the axis The forces may be acting on the axis of rotation The forces may be acting parallel to the axis of rotation The TORQUE caused by some forces may be equal and opposite to that caused by other forces Solution :For all the options, NET external torque on a system of particles can be ZERO. So, allare compatible with it.

Discussion

8.	Explain "By changing unit zero are added. This will not change number of significant digits.''
Answer» Solution : Let mass of a PEARL is 85 milligram which is measured upto accuracy of 1 milligram. If this mass is measured in TERMS of gram it will be represented as 0.085. In this number there will be 2 significant digit. Same mass is measured in terms of kilogram - 0,00 0085 kg. Here, also UNDERLINED zero are not significant digit. Therefore, no. of significant digit will be 2. This shows that by changing unit no. of significant digit will not CHANGE.

Discussion

9.	For a gram-molecule of an ideal gas (PV)/T is
Answer» 4.2 calories 8.3 calories 4.2*10^3 ergs 2 calories Answer :D

Discussion

10.	To increase the magnifying power of telescope (f_0= focal length of the objective and f_e= virtual length of the eye lens)
Answer» `f_0` should be LARGE and `f_e` should be SMALL `f_0` should be small and `f_e` should be large `f_0 and f_e` both should be large `f_0 and f_e` both should be small Answer :A

Discussion

11.	The two length scales (range in the study of scale of length) differ by a factor of
Answer» `10^20`or more `10^40` or more `10^28` or more `10^20` ANSWER :B

Discussion

12.	(a) Ice floats in water with about wine-tenths of its volume submerged. What is the fraction sub merged for an iceberg floating on freshwater lake of a hypothetical) planet whose gravity is to times that of the earth? (b) What is the fractional volume submerged of an ice cube in a fall of water placed in an enclosure which is freely falling under gravity?
Answer» Solution :(a) The fractional volume submerged does not depend on the gravity. So in the PLANET also fee will get submerged upto the same level. The CONDITION for flotation is: mass of floating body mass of LIQUID displaced (B) Pressure and hence UPTHRUST depends on g. So ifg=0upthrust is zero. So the ice will float with any value for the fractional submerged volume.

Discussion

13.	In the previous problem, the tension in the string is
Answer» `(MG)/(4)` `(mg)/(3)` `(mg)/(2)` `(2MG)/(3)` SOLUTION :`mg-T=ma` ` =(2mg)/(3)` `T=(mg)/(3)`

Discussion

14.	Pressure and stress having same unit (and dimension) although they are not a same physical quantity, why ?
Answer» SOLUTION :Pressure is a SCALAR QUANTITY while stress is a VECTOR quantity.

Discussion

15.	Find the speed of two objects such that when they move uniformly towards each other , they get 4.0 meters closer each second and when they move uniformly in the same direction with original speeds , they get 4.0 metres closer each 10 sec.
Answer» SOLUTION :Let their SPEEDS be `v_(1)"and " v_(2) ` and let `v_(1)gtv_(2)`. In First Case :Realative VELOCITY , `v_(1)-v_(2)=4 `….(1) In Sencond Case: Relative velocity ,`v_(1)-v_(2)=4/(10)=0.4`….(2) Solving eqns. (1)and (2), we GET `v_(1)=2.2ms^(-1), v_(2) = 1.8 ms^(-1)`.

Discussion

16.	Two rigid bodies have same angular momentum about their axis of rotation. Which of the bodies will have greater kinetic energy?
Answer» Solution :RELATION between ANGULAR MOMENTUM and kinetic energy is KE =`(L^(2))/(2I)` Even though angular momentum is same, the body with less moment of INERTIA will have greater kinetic energy.

Discussion

17.	During fall of water gently from a water-tap the stream gets thinner at the bottom. Why?
Answer» Solution :Velocity of downward flow is HIGHER. So the streamlines are CROWDED TOGETHER. HENCE the flow becomes thinner.

Discussion

18.	The dimensional formual for gravitational constnat G is
Answer» `[ML^(3)T^(-2)]` `[M^(-1)L^(3)T^(-2)]` `[M^(-1)L^(-3)T^(-2)]` `[ML^(-3)T^(2)]` Answer :C

Discussion

19.	A particle of mass m is moving yz-plane with uniform velocity v with its trajectory running parallel to +ve y - axis and intersecting z-axis at z=a The change in its angular momentum about the origin as it bounces elastically from a wall at y= constant is
Answer» MVA`OVERLINE(i)` 2 mva `overline(i)` `-"mva" overline(i)` `="2amv" overline(i)` ANSWER :B

Discussion

20.	A homogeneous chain of length 2l and mass M lies on an absolutely smooth table. A small part of the chain hangs from the table. At the initial moment, the part of the chain lying on the table is held and released, after which the chain begins to slide off the table under the weight of the hanging end. Find the velocity of the chain when the length of the hanging part is equal to x(x lt I). Also calculate the acceleration and the force with which it acts on the edge of the table.
Answer» ANSWER :`[xsqrt((g)/(2L)),(xg)/(2l),(SQRT(2)Mg)/(4l^(2))(2l-x)x]`

Discussion

21.	Two satellites A and B are revolving around the earth circular orbits of radius r_(1) and r_(2) respectively with r_(1) lt r_(2). Plane of motion of the two are same. At position 1, A is given an impulse in the direction of velocity by firing a rocket so that it follows an elliptical path to meet B at position 2 as shown. A?t position 2, A is given another impluse so that velocities of A and B at 2 become equal and the move together. For any elliptical path of the satellite of the time period of revolution is given by Kepler's planetary law as T^(2)alpha r^(3) where a is semi major axis of the ellipse which is (r_(1)+r_(2))/(2) in this case. Also angular mopmentum of any satellite revolving around the Earth will remain a constant about EArth's centre as force of gravity on the satellite which keeps it in elliptical path is along its position vector relative to the earth centre. When A is given its first impulse at the moment.
Answer» A, B and earth centre are in same STRAIGHT line B is AHEAD of A ANGULARLY B is behind A angular NONE of these Answer :B

Discussion

22.	The diagram shows stress v/s strain curve for the materials A and B. From the curves we infer that:
Answer» A is BRITTLE but B is DUCTILE A is ductile but B is brittle both a and B are ductile Both A and B are brittle Answer :2

Discussion

23.	A physical quantity X is related to three observable a,b and c as X=(b^(2)sqrt(a))/c. The errors of measuremnts in a,b and c are 4%, 3% and 2% respectively. What is the percentage error in the quality X?
Answer» SOLUTION :`R=V//I=11.1//5.5=2.02Omega, DELTAR//R=DeltaV//V+DeltaI//I=0.1//11.1+0.5//5.5=0.099` `DeltaR=(0.099xx11.1)//5.5=0.199=0.2Omega`.RESISTANCE `=(2.02+-0.2)OMEGA`

Discussion

24.	g is dimensional constant where as pi is the dimensionless constant. Why?
Answer» SOLUTION :g has the DIMENSIONLESS formula `[LT^(-2)],pi`has no DIMENSIONS.

Discussion

25.	A hemispherical bowl of radius (R) is set rotating about its axis of symmetry which is kept vertical. A small block kept in the bowl rotates with bowl without slipping on its surface. If the surface of bowl is smooth, the angle made by the radius through the block with the vertical is theta.the angular speed at which the bowl is rotating
Answer» `SQRT((G)/(rcostheta))` `sqrt((gcostheta)/(R))` `sqrt((gsintheta)/(R))` `sqrt((g)/(Rsintheta))` ANSWER :A

Discussion

26.	A uniform triangular plate is kept horizontal suspended with the help of three vertical threads as shown. The sides of the plate have length l_(1), l_(2) and l_(3). Find tension T_(1), T_(2) and T_(3) in the three threads. Mass of the plate is M.
Answer» ANSWER :`T_(1) = T_(2) = T_(3) = (MG)/(3)`

Discussion

27.	A satellite is orbitingaround the earth in an orbit in equatorial plane, of radius 2R_(e), where R_(e), is the radius of th earth. Findthe area on the earth, this satellite covers for communication purpose in its complete revolution
Answer» `4piR_(E)^(@)` `2pisqrt(3)R_(e)^(2)` `2pi(2-SQRT(3))R_(e)^(2)` `2pi(4 + sqrt(3))R_(e)^(2)` ANSWER :D

Discussion

28.	An equilarteral triangle ABC is formed by joining three rods of equal length and Dis the mid point of AB. The coefficient of linear expansion for AB is alpha_(1) and for AC and BC is alpha_(2), if distance DC remains constant for small changes in temper- ature. Find the relation between alpha_(1) and alpha_(2)(##AKS_NEO_CAO_PHY_XI_V01_PMH_C12_SLV_016_Q01.png" width="80%">
Answer» SOLUTION :D is MID point condition to have CD constant `AC^(2) - AD^(2) `constant `L_(2)^(2) - (L_(1)^(2))/(4)` = constant Differentiating w.r.tt `2L_(2)(dL_(2))/(dt ) - 2 (L_(1))/(4) (dL_(1))/(dt) = 0 ` or `2L_(2) dL_(2) - 2 (L_(1))/(4) dL_(1) = 0` `dL_(2) = L_(2) alpha_(2) Delta theta and dL_(1) = L_(1) alpha Delta theta` ` 2 L_(2) , L_(2) alpha_(2) Delta theta = (L_(1))/(2) L_(1) theta_(1) Delta theta` but `L_(2) = L_(1)` `therefore 2 alpha_(1) = (1)/(2) alpha_(1) "or " 4 alpha_(2) = alpha_(1)`

Discussion

29.	The bob A of a pendulum released from 30^(@) to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic.
Answer» SOLUTION :It transfers its entire MOMENTUM to the BALL on the table, and does not rise at all.

Discussion

30.	A body executes SHM such that its velocity at the mean positon is 1ms^(-1) and acceleation at the extreme position is 1.57ms^(-2). Calculate the amplitude and the time period of oscillation.
Answer» Solution :The velocity at the MEAN POSITION is MAXIMUM given by `v_("MAX")=Aomega=1ms^(-1)` ……….1 The acceleration at the extreme position is maximum given by `a_("max")=Aomega^(2)=1.57ms^(-2)`……….2 Dividing 2 by 1 We get `(a_("max"))/(v_("max"))=(Aomega^(2))/(Aomega)=1.57/1` or `omega=1.57` i.e.`(2pi)/T=1.57` rad `:.` Time period `T=(2pi)/1.57=(2(3.14))/1.574s` From Eq. (1) `A omega =1` i.e `A(1.57)=1` or `A=1/1.57` `:.` Amplitude `A=0.637m`

Discussion

31.	If m is the mass of drop of a liquid of radius 'r' then (mg)/(pi r) has the same dimensions of:
Answer» SURFACE tension tension young's modulus coeffici ENT of viscosity Answer :A

Discussion

32.	The kinetic energy is conserved in
Answer» ELASTIC collision inelastic collision both (a) and (B) none Answer :A

Discussion

33.	The length og two metal wire is l_(1) when the tension in it is F_(1) and l_(2) when the tension is F_(2) .then the original length of the wire is
Answer» Solution :Let I be the natural length of the wire. For the force` F_(1)` if elongation is `e_(1)` `e_(1)=(F_(1)l)/(AY) rArr Y=(F_(1)l)/(Ae_(1))` Similarly for the force `F_(2)`, if elongation is `e_(2)` `e_(2)=(F_(2)l)/(AY)= rArr Y=(F_(2)l)/(Ae_(2)), l_(2)=l+e_(1) and l_(2)=l+e_(2)` `Y=(F_(1)l)/(A(l_(1)-l))=(F_(2)l)/(A(l_(2)-l))= F_(1)(l_(1)-l)=F_(2)(l_(1)-l) and F_(1)l_(2)-F_(1)=F_(2)l_(1)-F_(2)l` `(F_(2)-F_(1))=l=l_(1)F_(2)-l_(2)F_(1) or l=(l_(1)F_(2)-l_(2)F_(1))/((F_(2)-F_(1)))`

Discussion

34.	The isothermal bulk modulus of a perfect gas at a atmospheric pressure is
Answer» `1.03xx10^(5) "NEWTON"//m^(2)` `1.03xx10^(6) "Newton"//m^(2)` `1.0370xx10^(10)" Newton"//m^(2)` `1.03xx10^(11)" Newton"//m^(2)` ANSWER :A

Discussion

35.	A body is projected vertically up with u. Its velocity at half its maximum height is
Answer» `U//2` `u^(2)//2` `SQRT(2)u` `u//sqrt(2)` ANSWER :D

Discussion

36.	If a simple pendulum is arranged in an artificial satellite its i) time period becomes infinity ii) frequency becomes infinity iii) both time period and frequency becomes infinity iv)it does not oscillate .
Answer» i and IV are correct i and ii are correct ii, iii are correct iii, iv are correct ANSWER :A

Discussion

37.	A rod AB of length l is pivoted at an end A and freely rotated in a horizontal plane at an angular speed omega about a vertical axis passing through A. If coefficient of linear expansion of material of rod is alpha, find the percentagechangein its angular velocity if temperature of system is increased by DeltaT.
Answer» Solution :If TEMPERATURE of surroundingincreases by `DeltaT`, the new lengthof rod becomes `l.=l(1+alphaDeltaT)` Due to change in LENGTH, moment of inertia of rod also changes and its is about and end A and is given as `I._(A)= (Ml^(2))/(3)` As no EXTERNALFORCE of torque is actingon rod, its angular momentumremainsconstant during HEATING. Thus we have `I_(A)omega=I_(A).omega.` [where `omega.` is the final anuglarvelocity of rod after heating] (or) `(Ml^(2))/(3)omega=(Ml^(2)(1+alpha DeltaT)^(2))/(3)omega.` `rArr omega=(1+2alpha DeltaT)omega^(1) or omega. = omega(1-2alphaDeltaT)` [using binomialexpansion for small `alpha`] Thus percentagechange in angularvelocity of rod due to heating can be given as `Delta omega=(omega-omega.)/(omega) xx 100%=2alpha DeltaT xx 100%`

Discussion

38.	A body starts from rest from a point distance R_(0) from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be (R rarr Radius of the earth) (R_(0) gt R)
Answer» `2GM((1)/(R) - (1)/(R_(0)))` `sqrt(2GM((1)/(R)-(1)/(R_(0))))` `GM((1)/(R)-(1)/(R_(0)))` `2GM sqrt(((1)/(R)-(1)/(R_(0))))` Answer :B

Discussion

39.	An external force acts on a rigid body, but the torque is zerp. What conclusions can be made from this ?
Answer» Solution :EITHER the PERPENDICULAR distance `vecr`from the axis of rotation to the LINE of action is ZERO or there is an EQUAL and opposite force.

Discussion

40.	A copper block of mass 500 gm and specific heat 0.1 cal/gm""^(@)C is heated from 30^(@)C to 290^(@)C, the thermal capacity of the block is
Answer» 50cal/`""^(@)C` 50GM 50cal/`""^(@)C` SGM ANSWER :A

Discussion

41.	A planet is at an average distance d from the sun, and its average surface temperature is T. Assume that the planet receives energy only from the sun, and loses energy only through radiation from its surface. Neglect atmospheric effects. If T prop d^(-n), the value of n is
Answer» 2 1 `(1)/(2)` `(1)/(4)` ANSWER :C

Discussion

42.	The comparison of any physical quantity with its standard unit is known as .........
Answer» FUNDAMENTAL QUANTITIES MEASUREMENT dualism DERIVED quantities Solution :measurement

Discussion

43.	In slipping, the resultant velocity of a point of contact acts along
Answer» FORWARD direction backward direction either (a) or (B) TANGENTIAL direction Answer :B

Discussion

44.	The time period of oscillation of the particle in SHM is 'T' Then match the following
Answer» a-e, B-g, c-h, d-f a-h, b-h, c-e, d-g a-f, b-e, c-h, d-g a-e, b-f, c-g, d-h Answer :B

Discussion

45.	A drop of some liquid of volume 0.04 cm^3 is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area 20cm^2 between the surface of the two slides. To separate the slides , force of 16xx10^5 dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyn. cm^-1)
Answer» 60 70 80 90 Solution :Thickness of layer, `h=("VOLUME")/("area")=(0.04)/(20)=0.002 cm` RADIUS of curvature of concave SURFACE, `r=(h)/(2)=0.001 cm.` Force, `F=T/rxx`area `THEREFORE` Surface tension. `T=(Fxxr)/("area")=(16xx10^5xx0.001)/(20) DYN. cm^-1`

Discussion

46.	Moment of inertia plays the same role in rotational motion as mass plays in translational motion' explain the statement.
Answer» Solution :When a force acts on a body some linear acceleration is produced in that body. Similarly ANGULAR acceleration is generated in a body due to APPLICATION of a torque on it. So the rotational ANALOGUES of force and linear acceleration are torque and angular acceleration RESPECTIVELY. Again in case oif linear motion force = mass `xx` acceleration and in case of circular motion torque = moment of inertia`xx` angular acceleration. Comparing THSES two relations it can be inferred that the rotational analogue of mass is themoment of inertia. So, moment of inertia in rotational motion plays the same role as mass plays in linear motion.

Discussion

47.	A sound travelling along a string is described by y = 0.005 sin (80.0x - 3.0t) calculate amplitude ,
Answer» SOLUTION :AMPLITUDE A = 0.005m

Discussion

48.	A system of two particles is having masses m_(1) and m_(2)If the particle of mass m, is pushed towards the center of mass of particles through a distance d. by what distance the particle of mass m_(2) should be moved so as to keep the centre of mass of particles at the original position?
Answer» `(m_(1))/(m_(1)+m_(2))d` d `(m_(1))/(m_(2))d` `(m_(2))/(m_(1))d` ANSWER :C

Discussion

49.	When a particle is performing SHM of time period T_1 the force acting on it is F_1 for a certain displacement. When the same particle is performing SHM of time period T_2 the force acting is F_1 for the same displacement. What will be the time period of the particle when a combined force of F_1 and F_2 produces the same displacement in SHM?
Answer» SOLUTION :`F_1 = K_1 X , F_2 = - K_2 x, F_1 + F_2 =- KX impliesK = K_1 + K_2` `m omega^2 = m omega_1^2 + m omega_2^2` `implies(1)/(T_2) = 1/(T_1^2) + 1/(T_2^2) ` (or) `T^2 = (T_1^2 T_2^2)/(T_1^2 + T_2^2) and T= (T_1 T_2)/(sqrt(T_1^2 + T_2^2))`

Discussion

50.	Resultant of two coplanar vector also will be planer vector.
Answer» ANSWER :TRUE.

Discussion

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The bob A of a pendulum released from 30^(@) to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic.

A body executes SHM such that its velocity at the mean positon is 1ms^(-1) and acceleation at the extreme position is 1.57ms^(-2). Calculate the amplitude and the time period of oscillation.

If m is the mass of drop of a liquid of radius 'r' then (mg)/(pi r) has the same dimensions of:

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A body starts from rest from a point distance R_(0) from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be (R rarr Radius of the earth) (R_(0) gt R)

An external force acts on a rigid body, but the torque is zerp. What conclusions can be made from this ?

A copper block of mass 500 gm and specific heat 0.1 cal/gm""^(@)C is heated from 30^(@)C to 290^(@)C, the thermal capacity of the block is

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Resultant of two coplanar vector also will be planer vector.