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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The periodic time of SHM also measured by t= 2pi sqrt(("displacement")/("acceleration")) can we say that the time period depend on the displacement? |
| Answer» Solution :No, because ACCELERATION also DEPEND on the displacement `(a= -omega^(2) x)`. | |
| 2. |
A man and a cart move towards each other. The man weighs 64 kg and the cart weighs 32kg. The velocity of the man is 5.4 km/hr and that of the cart is 1.8 km/hr. When the man approaches the cart, he jumps on to it. The velocity of the cart carrying the man will be |
| Answer» Solution :`m_(1)u_(1)+m_(2)u_(2)=(m_(1)+m_(2))V` | |
| 3. |
The net rate of heat loss by a hot body depends upon |
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Answer» TEMPERATURE of body |
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| 4. |
Which of following statements is incorrect? |
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Answer» Path LENGTH is a scalar quantity whereas displacement is a vector quantity |
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| 6. |
9765 div 14.39 = 678.59625. This quotient on rounding off to two significant figures is |
| Answer» ANSWER :A | |
| 7. |
A container si filled up to its brim with water with a piece of ice floating on it [Fig 5.15(a)]. State the effect on the level of water in it container when ice melts completely, if the initial temperature of water was (i) 0^(@)C (ii) 4^(@)C and (iii) above 4^(@)C. |
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Answer» Solution :VOLUME expansion of the container at the temperatures mentioned is neglected. (i) There will be no change in the water level. Here mass of ice is equal to the mass of the displaced water as per the condition of floatation. Ice, being at `0^(@)C`, produces water at `0^(@)C` on melting. Volume of that water is equal to the space formerly occupied by the immersed portion of ice. Hence, water THUS formed on melting of ice exactly fills the space [Fig.5.15(b)].  (II) Some water will OVERFLOW from this container. Density of water is maximum at `4^(@)C`, so water formed on melting of ice at `0^(@)C` is of lower density. Therefore the volume of this water will be greater than the volume of water displaced by floating ice at `4^(@)C` [Fig 5.15(c)]. (iii) The water level will come down [Fig.5.15(d)]. Mass of the ice floating is equal to the mass of water displaced by the ice. According to the question, temperature of the water is above `4^(@)C`. Receiving heat from the surroudings and water, the ice produces water at `0^(@)C` on melting. Density of water at `0^(@)C` is higher than that of water at `4^(@)C`.Hence, volume of the water displaced is more than the volume of the water produced on melting. Also, temperature of the water falls as it supplies heat to the ice. With decrease in temperature, volume of the water also DECREASES. |
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| 8. |
In an experiment, the angle are required to be measured using an instrument. 29 divisions of the main scale coincide with 30divisions of the vernier scale. If the smallest division on the main scale is half a degree (0.5^(0)), then the least count of the instument is |
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Answer» HALF MINUTE |
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| 9. |
A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30^(@). The co-efficient of static friction mu_(k)=0.25. If the inclination theta of the plane is increased, at what value of theta does the cylinder begin to skid, and not roll perfectly ? |
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Answer» `37^(@)` |
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| 10. |
If a glass prism is dipped in water, its dispersive power |
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Answer» increases |
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| 11. |
A steel ball is dropped in oil then the ball attains _________ after some time. |
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Answer» CONSTANT force |
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| 12. |
A light stright fixed at one end to a wooden clamp on a ground passes over a fixed pulley and hangs at the other side. It makes an angle 30^@with the ground. A clamp can tolerate upto vertical force 360N. Find the, maximum acceleration in the upward direction with which the boy can climb safely. The friction of pulley and wooden clamp may be ignored |
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Answer» `1m//s^2` |
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| 13. |
A half metre scale is pivoted about a horizontal fiictionless pin through one end as shown in Fig. The scale is released from rest in a vertical position. At the instant when the scale is horizontal what is (i) its angular speed and (ii) the magnitude of the angular acceleration.(Given length of rod =1/2 m) |
| Answer» Solution :`mg(l//2) = (1//2) IOMEGA^(2), I=ml^(2)//3, l=1//2 , omega = sqrt(6G), ALPHA = tau//l = mg xx (1//4) (m//12) = 3g` | |
| 14. |
A pendulum of length 60 cm is suspended inside an aeroplane. The aeroplane is flying up with an acceleration of 4m*s^(-2) making an angle of 30^(@) with the horizontal. Find the time period of oscillation of the pendulum. |
Answer» Solution :Horizontal component of ACCELERATION a of the plane = `acos30^(@)` and vertical component = `acos30^(@)`  Hence, the downward acceleration experienced by the pendulum BOB = `g-(-asin30^(@))=g+asin30^(@)` So, the acceleration of the pendulum bob `g.=sqrt((g+asin30^(@))^(2)+(acos30^(@))^(2))` `=sqrt((9.8+4xx1/2)^(2)+(4xxsqrt3/2)^(2))=sqrt((11.8)^(2)+12)` `=sqrt(151.24)=12.3m*s^(-2)` `therefore` TIME period of the pendulum `T=2pisqrt(L/(g.))=2xxpisqrt(0.60/12.3)=1.38s`. |
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| 15. |
A metallic ball of radius 0.2 m is heated from 320 K to 522 K. What will be the increase in surface area of the ball?Given, coefficient of superficial expansion of the ball = 3.4 xx 10^(-5)K^(-1) |
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Answer» Solution :Given , RADIUS of ball at 320 K , `r_320 = 0.2 m` `DELTA T = 522 - 320 = 202 K` ` beta = 3.4 xx 10^(-5) K^(-1)` surface area of ball at 320 K, `S_320 = 4pi r_(320)^2` `= 4pi (0.2)^2 = (16pi)/(100) m^2` increase in surface area is `Delta S = S_(320) beta Detla T = (16pi)/(100) xx 3.4 xx 10^(-5) xx 202` ` = 3.452 xx 10^(-5) m^2` |
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| 16. |
A particle is moving such that its position vector varies with time as vec(r) = (1 - alpha t) t vec(A) , where alpha and vec(A) are constant quantities. At t = 0, the particle is at a position O. At some later instant 't_(0)', the particle is again at O. Velocity of the particle at the instant t_(0) is: |
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Answer» `2 vec(A)` `(d vec(r))/(DT) = (1- 2 alpha t) vec(A)` |
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| 17. |
An iron sphere of mass 20xx10^-3kgfalls through a viscous liquid with terminal velocity 0.5 ms^(-1) The terminal velocity (in ms^-1) of another iron sphere of mass 54 xx 10^2 kg is |
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Answer» 4.5 |
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| 18. |
A simple pendulum of time period 1s and length l is hung from a fixed support at 0. Such that the bob is at a distance H vertically above A on the ground (figure) the amplitude is theta_(0) the string snaps at theta=(theta_0)/(2). Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume theta_(0) to be small, so that sin theta_(0) approx theta_(0)" and"cos theta_(0) approx 1. . |
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Answer» Solution :Two components of velocity Perpendicular component to the string `v_(y)= l omega sin ""(theta_0)/(2)` Prallel component to the string `v_(x)= l omega cos (theta_0)/(2)`  From `V= r omega` `therefore v= l omega` SUPPOSE at `t=0, THETA = theta_0` `therefore theta = theta_(0) cos omega t` but `T= 1s, therefore omega (2pi)/(T)= 2pi rad"/"s` `therefore theta= theta_(0) cos 2pi t"""........"(1)` Now at `t= t_(1)` time, `theta= (theta_0)/(2)` `therefore (theta_0)/(2)= theta_(0) cos 2pi t_(1)` `therefore (1)/(2) = cos 2pi t_(1)` `therefore 2pi t_(1) = (pi)/(3)` `therefore t_(1)= (1)/(6)s` Difference equation (1) with time, `(d theta)/(dt)= -theta_(0) 2pi sin 2pi t` `therefore t= t_(1)` time `omega = -2pi theta_(0) xx sin 2pi xx (1)/(6)` `=-2pi theta_(0) xx sin ""(pi)/(3)` `= -2pi theta_(0) xx sin"" (sqrt(3))/(3)` `therefore omega = -sqrt(3) pi theta_(0)` (-)ve shows that bob.s motion is towards, left. Since linear velocity is in the tangential direction. `therefore v= l omega""[therefore v= A omega]` `therefore v= -sqrt(3) pi theta_(0) l` The Y-component of linear velocity `v_(y)= -sqrt(3) pi theta_(0) l sin ""(theta_0)/(2)` and x-component of velocity `v_(x)= -sqrt(3) pi theta_(0) l cos"" (theta_0)/(2)` When string break, the height of bob from the ground, `H. = H+l(1- cos ""(theta_0)/(2))"""........"(2)` Time taken by the bob from this height (when string break) to ground it t, then `H.= v_(y)t+(1)/(2)G t^(2)` (all terms are (-)ve) `therefore (1)/(2) g t^(2) + sqrt(2) pi theta_(0) l sin ""(theta_0)/(2) l-H.=0` is a QUADRATIC formula of t `therefore triangle = b^(2) -4ac` `=3p^(2)theta_(0)^(2)sin^(2) ""(theta_0)/(2)+4xx(1)/(2)gxx H.` `=3pi^(2) theta_(0)^(2) l^(2) sin^(2) ""(theta_0)/(2) + 2gH.` but `theta_0` is small `therefore sin^(2) ""(theta_0)/(2)=0`, `triangle = 2gH.` `therefore sqrt(triangle) = sqrt(2gH.)` `therefore t= (-b pm sqrt(triangle))/(2a)` `= (-sqrt(3) pi theta_(0) l sin ""(theta_0)/(2) pm sqrt(2gH.))/(2xx (1)/(2)g)` but `sin""(theta_0)/(2)=0`, `t= (pm sqrt(2gH.))/(g)` `therefore t= sqrt((2H.)/(g))"""........"(3)` Now, in `H. = H+l(1-cos ""(theta_0)/(2)), theta_(0)` is small `cos ""(theta_2)/(2) approx 1` `therefore H.= H+l(1-1)` `therefore H.= H` `therefore t= sqrt((2H)/(g))""[" from equation (3) "]"""........."(4)` The bob covered distance in horizontal direction. `X= v_(x) t` `= -sqrt(3) pi theta_(0) l cos ""(theta_0)/(2)xx t` Since `theta_(0)` is small, `cos ""(theta_0)/(2) approx 1` and taking `t= sqrt((2H)/(g))`, `X= -sqrt(3) pi theta_(0) lxx sqrt((2H)/(g))` `therefore X= -sqrt((6H)/(g))xx pi theta_(0) l"""........"(5)` but string breaks, the bob is at RS distance from vertical at that time and `RS= l sin ""(theta_0)/(2)` `RS= (l theta_0)/(2)""[therefore ""sin (theta_0)/(2) approx (theta_0)/(2)]` When bob hit at ground, the distance from bob at that time `=(l theta_0)/(2)+X` `= (l theta_0)/(2) - sqrt((6H)/(g)) pi theta_(0) l` `= theta_(0) l((1)/(2)-pi sqrt((6H)/(g)))`. |
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| 19. |
A rectangular block of dimensions 6m xx 4m xx 2m and of density 1.5 gm/c.c is lying on horizontal ground with the face of largest area in contact with the ground. The work done in arranging it with its smallest area in contact with the ground is. (g=10 ms^(-2)) |
| Answer» Answer :B | |
| 20. |
A large bottle is fitted with a siphon made of capillary glass tubing. Compare the times taken to empty the bottle when it is filled. (i) with water (ii) with petrol of density 0.8 cgs units. Theviscosity of water and petrol are 0.1 and 0.2 cgs units repectively. |
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Answer» Solution :The volume of liquid flowing in time t through a capillary tube is given by `V=Qt=(pi Pr^(4)t)/(8 eta l)=(pi h rho gr^(4)t)/(8 etal)` `:.` For Water `V_(1)=(pi h rho_(1)gr^(4)t_(1))/(8l eta_(1))` For PETROL , `V_(2)=(pi h rho_(2)gr^(4)t_(2))/(8l eta_(2))` But `"" V_(1)=V_(2)` `:. "" (pi h rho_gr^(4)t_(1))/(8l eta_(1))=(pi h rho_(2)gr^(4)t_(2))/(8l eta_(2))` `or ""(t_(1))/(t_(2))=(eta_(1))/(eta_(2))XX (rho_(1))/(rho_(2))=(0.01)/(0.02)xx(0.8)/(0.02)xx(0.8)/(0.02)=0.4` |
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| 21. |
Two perfectly elastic spheres of masses 2kg and 3kg moving in opposite directions with velocities 8 ms^(-1) and 6 ms^(-1) respectively collide with each other. Find their velocities after the impact |
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Answer» Solution :Here `m_(1)= 2KG, m_(2)= 3kg` `u_(1)= 8ms^(-1), u_(2)= -6 ms^(-1)` According to the law of conservation of LINEAR momentum `m_(1)u_(1) + m_(2)u_(2)- m_(1)v_(1) + m_(2)v_(2)` i.e., `(2)(8) + (3) (-6) = 2v_(1) + 3v_(2)` `rArr 2v_(1) + 3v_(2) = -2` ....(i) SINCE the collision is elastic, `u_(1)-u_(2)= v_(2)-v_(1)` i.e., `8-(-6)= v_(2)-v_(1)` `rArr v_(2)-v_(1)= 14` ...(ii) On SOLVING the equations (i) and (ii) we get `v_(1)= -8.8 ms^(-1), v_(2)= 5.2 ms^(-1)` |
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| 22. |
While deriving the expression of the pressure of a ideal gas, |
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Answer» only TOTAL KE remains conserved |
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| 23. |
(A) : Kerosene Oil spreads out over water surface. (R ) : Surface tension is a surface phenomenon. |
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Answer» Both (A) and (R ) are true and (R ) is the correct explanation of (A) |
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| 24. |
A man carries a bucket containing water in one hand and a live fish in the other. If he releases the fish into the bucket of water, will he carry less weight? |
| Answer» Solution :The man will CARRY the same weight when he releases the live fish into the bucket of water. When the fish is released into the water of the bucket, the displaced water exerts an upthrust and hence the apparent weight of the fish will decrease. According to Newton.s third law of motion, an EQUAL reaction force will ACT downwards at the bottom of the bucket. So, the decrease in the weight of the fish becomes equal to the increase in the THRUST at the bottom of the bucket. As a result, the net weight of the bucket will remain the same and hence the man will carry the same weight as before. | |
| 25. |
It work done positive or negative for stretched spring ? |
| Answer» SOLUTION :POSITIVE .Because PULLING and DISPALCEMENT are in the same DIRECTION . | |
| 26. |
The coefficient of friction between a plane inclined at 60^(@) and a body of mass 100 g kept on that plane is (1)/(sqrt(3)),. What minimum force should be applied along the plane so that the body can move upwards ? |
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Answer» |
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| 27. |
What is the value of tangential component of a linear acceleration for a representation particle moving in uniform angular motion rotational motion? |
| Answer» Solution :The tangential component of a LINEAR ACCELERATION `a_(T)` for a particle moving with CONSTANT ANGULAR velocity in rotational motion is `a_(T)=ralpha,` here `alpha=0. THEREFORE a_(T)=0` | |
| 28. |
Two cliff of heights 120m and 100.4m are separated by a horizontal distance of 16m. If a car has to reach from the first cliff to the second , the horizontal velocity of the car should be |
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Answer» `16 MS^(-1)` |
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| 29. |
Two bodies having the same mass 5kg each have different surface areas 20m^(2) and 10m^(2) in contact with a horizontal plane. If the coefficient of friction is 0.4, the forces of friction that come into play when they are in motion will be in the ratio |
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Answer» `1:1` |
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| 30. |
When 1 gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphere, thevolume increases from 1cc to 1671cc. The heat of vaporisation at this pressure is 540 cal/gm. Increase in internal energy of water is |
| Answer» Answer :D | |
| 31. |
Two blocks of masses 10 kg and 30 kg are placed along a vertical line. The first block is raised through a height of 7 cm. By what distance should the second mass be moved to raise the centre of mass by 1 cm ? |
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Answer» 1 CM upward |
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| 32. |
Three moles of an ideal gas (C_p = 7/2 R) at pressure P_0 and temperature T_0 is isothermally expanded to twice its initial volume, it is then compressed at a constant pressure to its original volume. Calculate net heat supplied to the gas during complete process (write your answer in terms of gas |
| Answer» SOLUTION :`3RT_0 In (2) -(21)/(4) RT_0` | |
| 33. |
The time period of a loaded spring on earth is 6s. On the surface of moon, the time period of the same loaded spring will be |
| Answer» ANSWER :D | |
| 34. |
A body executes SHM such that its velocity at mzean position is 2m/s and acceleration at extreme position is 4m//s^2 Find the time period of the motion. |
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Answer» Solution :At MEAN POSITION, velocity is MAXIMUM i.e., A`omega`= 2. Similarly at extreme position acceleration is maximum i.e. `A omega^2 = 4pi` from the above TWO relatives `w= 2pi ` c/s Time PERIOD `T= (2pi)/(omega) = (2pi)/(2pi)= 1s` |
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| 35. |
A steel tape measures the length of a copper rod as 80 cm at 15^(0)C, which is the calibration temperaturefor the tape. Find reading of the tape when both are at a temperature of 65^(0)C. (alpha of steel and copper are 12xx10^(-6)//""^(0)C and 17xx10^(-6)//""^(0)C) |
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Answer» |
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| 36. |
A bottle with an open mouth contains air at 47°C. The temperature to which the bottle must be heated that 1/5th of the mass of air leaves the bottle is _______ °C |
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Answer» 172 |
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| 37. |
Should the centre of mass of a body necessarily lie inside the body? |
| Answer» SOLUTION :Not necessarily. For example the CENTRE FO mass of L shaped ROD lies in the region outside the rod. | |
| 38. |
Statement I : A particle performs a simple harmonic motion with amplitude A and angular frequency omega. To change the angular frequency of the simple harmonic motion to 3omega and amplitude to A/2, we have to supply an extra energy of (5/4)momega^(2)A^(2), where m is the mass of the particle executing simple harmonic motion. Statement II : Angular frequency of simple harmonic motion is independent of amplitude of oscillation. |
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Answer» Statement I is true, statement II is true , statement II is a CORRECT EXPLANATION for statement I. |
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| 39. |
A block of mass m is relesed from rest from a height h onto a smooth sledge of mass M fitted with an ideal spring of stiffness k. |
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Answer» SOLUTION :Method 1: a. As no external force in horizontal direction, the LINEAR momentum will be conserved in horizonta direction. Let thevelocity of block and wedge be a and `V` respectively up to the case when the block reaches onhorizontal surface on the wedge. `mv=MV` ...............i Mechanical energy is also conserved in this case `/_\K+/_\U=0` `(1/2mv^(2)+1/2MV^(2)-0)+(-mgh)=0` `1/2mv^(2)+1/2M(m/Mv)^(2)=mgh` `v=SQRT((2Mgh)/(M+m))` and `V=sqrt((2m^(2)gh)/(M(M+m)))` Method 2 : Applying work energy theorem from centre of mass frame `W_("gravity")=(/_\K)_(CM)` `mgh=1/2((mM)/(m+M))(v+V)^(2)` also, `mv=MVimpliesv=(MV)/m` `impliesmgh=12((mM)/(m+M))V^(2)(M/m1)^(2)` `implies V^(2)=sqrt((2m^(2)gh)/(M(m+M)))` b. INITIAL linear mometum to of the system is zero and there is no external force on the system., hence at the time of maximum compression of spring the linear momentum of the system should be zero. Hence final kinetic energy should be zero. `/_\U+/_\K=0implies/_\K=K_(f)-K_(i)=0` Hence, `/_\U_("gravity")+/_\U_("spring")=0` `-mgh+1/2kx^(2)=0impliesx=sqrt((2mgh)/k)` |
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| 40. |
Three moles of an ideal gas (C_p = 7/2 R) at pressure P_0 and temperature T_0 is isothermally expanded to twice its initial volume, it is then compressed at a constant pressure to its original volume. Calculate net work done by the gas |
| Answer» SOLUTION :`3RT_0 In (2) -3/2RT_0` | |
| 41. |
A body of mass 50 kg and specific gravity force is to be lifted using a rope of breaking strength of 60 kg wt. The acceleration with which it can be pulled up in least interval of time |
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Answer» ZERO |
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| 42. |
Action and reaction forces acts on |
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Answer» Same body |
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| 43. |
State the factors on which the work done by the force depends on. |
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Answer» Solution :(i) MAGNITUDE of the force. (ii) Magnitude of the DISPLACEMENT of the body. (iii) ANGLE between the force and displacement. |
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| 44. |
Two blocks A and B are connected to each other by a string and a spring. The string passesover a frictionless pulley as shown in the figure. Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of blocks is 0.2. Force constant of the spring is 1960 N/m. If mass of block A is 2 kg. Calculate the mass of block B and the energy stored in the spring. |
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Answer» |
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| 45. |
A flask contains 100 c.c of a liquid at 10^(0) C. When it is heated to 110^(0)Cincrease in volume of the liquid appears to be 2 c.c. Find the coefficient of real expansion of the liquid. (alpha of flask is 11 xx 10^(-6)//^(0) C ). |
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Answer» Solution :Coefficient of apparent expansion `gamma_(a) = (v_(2) - v_(1))/(v_(1) (t_(2) - t_(1))) = (2)/(100 (110 - 10)) = 2 xx 10^(-4) //^(0) C ` Coefficient of cubical expansion of the FLASK`(gamma) = 3 ALPHA` `= 3 xx 11 xx 10^(-6) = 33 xx 10^(-6)//^(0)C ` Coefficient of real expansion of the liquid, `gamma_(R) = gamma_(a) + gamma` `= 2 xx 10^(-4) + 33 xx 10^(-6) = 2 xx 10^(-4) + 0.33 + 10^(-4)` `therefore gamma_(R) = 2.33 xx 10^(-4) //^(@) C ` |
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| 46. |
A small steel ball falls through a syrup at a constant speed of 1.0m/s. If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upwards? |
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Answer» Solution :`W_(e)` = EFFECTIVE weight In equilibrium, `kv=w_(e)` Again, `2w_(e)-w_(e)=kv^(1)` From Eqs. (i) and (II) `v^(1)=v=1.0m//s` 
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| 47. |
The workdone by a force (2hat(i) - hat(j) + 5hat(k)) when it displaces the body from a point (3,4,-6) to a point (7,2,5) is |
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Answer» 65 units |
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| 48. |
A drop of liquid of density rho is floating half-immersed in a liquid of density d. If sigmais the surface tension, the diameter of the drop of the liquid is : |
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Answer» `SQRT((SIGMA)/(g (2 RHO -d))` |
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| 49. |
Can a single isoloated force exist in nature explain your answer |
| Answer» SOLUTION :Noaccordingto newton'sIII lawforcesalwaysexistin PAIRS . | |
| 50. |
In a resonance tube experiment to determine the speed of sound in air , a pipe of diameter5 cm is used . The air column in pipe resonates with a tuning fork of frequency 480 Hz , when minimum length of air column in 16 cm . Find the speed of sound in air at room temperature. |
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Answer» Solution :Fundamental FREQUENCY of air column closed at ONE end is `n = (v)/( 4(l + e)) = (v)/( 4(l + 0.3 D))` GIVEN ` n = 480 Hz , D = 5 CM = 5 xx 10^(-2) m` `l = 16 cm = 16 xx 10^(-2) m` `v = 4 n( l + 0.3 D)` `= 4 xx 480 [ 16 xx 10^(-2) + 0.3 xx 5 xx 10^(-2)] m//s` `= 4 xx 480 xx 17.5 xx 10^(-2)m//s = 336 m//s` |
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