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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.
| 14001. |
A body of mass m collides elastically with a stationary body of mass M. After collision, m has a speed equal to one-third of its initial speed. Calculate the ratio (m/M). |
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Answer» SOLUTION :`v_1 = ((m_1 - m_2)u_1)/((m_1 + m_2)) ` when ` u_2 = 0` `:. (u_1)/(3) = ((m - M)u_1)/((m + M))` `:. 3M - 3M = m + M or 2m = 4M` or `m/M = 4/2 = 2/1`. |
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| 14002. |
In a horizontal pipe of non-uniform cross section, water flows with a velocity of 1ms^(-1) at a point where the diameter of the pipe is 20 cm. The velocity of water (ms^(-1)) at a point where the diameter of the pipe 10 cm is: |
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Answer» 0.0025 m/s `d_(1)=20xx10^(-2)m` `v_(2)=?` `d_(2)=10xx10^(-2)m` `(5)/(4)d_(1)^(2)v_(1)=(5)/(4)d_(2)^(2)v_(2)` `d_(1)^(2)v_(1)=d_(2)^(2)v_(2)` `(v_1)/(v_2)=(d_(2)^(2))/(d_(1)^(2))` `(1)/(v_2)=((20xx10^(-2))^(2))/((10xx10^(-2))^(2))` `(1)/(v_2)=4` `v_(2)=(1)/(4)m//s` `=.25m//s` |
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| 14003. |
A particle executes SHM along X-axis. The force acting on it is given by………. |
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Answer» `-AKX` Force constant `k= AK` `F= -Akx`. |
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| 14004. |
Statement I: The uplift of the wing of an aircraft moving horizontally is caused by a pressure difference between the upper and lower faces of the wing. Statement II: The velocity of air moving along the upper surface is higher than that along the lower surface. |
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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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| 14005. |
For a wave described by y = 0.5 sin(x - 60t) cm, find (i) amplitude (ii) wave-vector (iii) wavelength (iv) angular frequency and frequency (v) periodic time (vi) speed of wave. |
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Answer» Solution :Comparing `y =0.5 sin (x-60t)` with `y = 0.5 sin (kx - omega t )` (i) amplitude of a wave `A =0.5 cm` (II) wave vector K =1 rad/cm (iii) wavelength `LAMDA = (2pi)/(k) = (2XX 3.14)/(1) = 6.28 cm` (iv) anguylar frequency `omega = 2pi f = 60` rad/s `therefore` Frequency `f = (60)/(2pi) = (60)/(2 xx 3.14)= 9.554 Hz` (v) PERIODIC time `T = (1)/(f) = (1)/(9.554) = 0.1047 s ` (vi) wave speed `v = (omeg )/(k) = (60)/(1) = 60 (cm)/(s)` |
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| 14006. |
An explosion blows a rock into three parts. Two pieces go off at right angles to each other, 1.0 kg piece with velocity of 12 m//s and other , 1.0 kg piece with a velocity of12 m//s and other 2.0 kg piece with a velocity of 8 m//s. If the third piece flies off with a velocity of 40 m//s compute the mass of the third piece. |
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Answer» Solution :Let `m_(1),m_(2)` and `m_(3)` be the MASSES of the three pieces. `m_(1)=1.0 kg, m_2=2.0 kg` Let `v_(1)=12 m//s, v_(2)=8 m//s, v_(3)=40 m//s`. Let `v_(1)` and `v_(2)` be directed along `x`- and `y`-axis respectively, and `v_(3)` be directed as shown.  By the principle of coservation of momentum, INITIAL momentum is zero. Hence, along `x`-axis, `0=m_(1)v_(1)-m_(3)v_(3)costheta` along `y`-axis `0=m_(2)v_(2)-m_(3)v_(3) sin theta` `implies m_(1)v_(1)=m_(3)v_(3) costheta` and `m_(2)v_(2)=m_(3)v_(3) sintheta` By squaring and adding we get `m_(1)^(2)v_(1)^(2)+m_(2)^(2)v_(2)^(2)=m_(3)^(2)v_(3)^(2)` `impliesm_(3)^(2)=(1^(2)(12)^(2)+(2)^(2)(8)^(2))/((40)^(2))impliesm_(3)=0.5kg` |
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| 14007. |
Which of the following statement are true for a stationary wave? |
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Answer» Every PARTICLE has a forced amplitude which is different from the amplitude of its nearest particle. |
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| 14008. |
A particle executes simple harmonic motion with an amplitude of 10cm and time period 6s , . At t = 0it is at position x = 5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4 s. |
| Answer» SOLUTION :x= (10 cm) SIN`((2pi)/(6) t + pi/6), ~~ 11 cm//s^2` | |
| 14009. |
What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20^@ C) is 2.50xx10^-2 N. m^-1?If an air bubble of the same dimension were formed at a depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 standard atmosphere pressure is 1.01xx10^5 Pa.). |
| Answer» Solution :Excess PRESSURE = 20.0 Pa (soap BUBBLE), 10.0 Pa (air bubble), TOTAL pressure `1.06xx10^5 Pa` | |
| 14010. |
A box of mass 50 kg at rest is pulled up on an inclined plane 12 m long and 2m high by a constant force of 100 N. When it reaches the top of the inclined plane if its velocity is 2ms^(-1), the work done against friction in Joules is (g=10 ms^(-2)) |
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Answer» 50 |
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| 14011. |
Define coefficient of static friction. An object of weight W rests on an inclined plane. The coefficient of friction between the object and the plane is mu. For what value of the angle of inclination theta, will the object move downward with uniform speed under its own weight ? |
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Answer» Solution :The forces ACTING on the body along with their COMPONENTS are shows in FIG. When the body is about to slide down along the INCLINE then, W sin `theta=mu R = mu Wcostheta` or, `tantheta=mu """or",theta= tan^(-1)mu` 
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| 14012. |
A stream of particles of mass m and separation d hits a perpendicular wall with a velocity v_(0) and rebounds along the original line of motion with a velocity v. The mass per unit length of the incident stream is lambda = (m)/(d). The the force exerted by the stream on the wall is |
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Answer» `LAMBDA V (v_(0) + v)` |
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| 14013. |
In the equation for I law of thermodynamics dQ=du+dw a) dQ depends on path b) dw depends on path c) du depends on path |
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Answer» only B is CORRECT |
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| 14014. |
A particle undergoes uniform circular motion. The velocity and angular velocity of the particle at an instant of time is v=3hat i+4hatj m//s and vec(omega )= xhat i+6hatjrad//sec The value of x in rad/s is |
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Answer» 8 |
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| 14015. |
What will be the velocity and acceleration of ball upwards at maximum height ? |
| Answer» Solution :VELOCITY is ZERO and acceleration is equal to .G. in downward DIRECTION. | |
| 14016. |
The relation between internal energy U, pressure P and volume V of a gas in an adiabatic process is U=a+bPV where a and b are positive constants. What is the value of gamma? |
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Answer» `a/B` |
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| 14017. |
An underformed spring of spring constant k is connected to a bead of mass m as shown in figure. Bead can move along long rigid rod without friction. A particle of mass moving with velocity v in the vertical plane containing spring and rod strikes bead at an angle 45^(@) with horizontal and sticks to bead. Choose correct alternative representing maximum elongation of spring is, all particles are in same horizontal plane. |
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Answer» `(SQRT(m/k))V` |
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| 14018. |
The period of oscillation of a harmonic oscillator isT = 2pisqrt("displacement/accelerations")Does T depend on displacement ? |
| Answer» SOLUTION :No. Because ACCELERATION ALSO DEPENDS on DISPLACEMENT | |
| 14019. |
If v_(1) and v_(2) be the velocities at the end of focal chord of projectile path and u is the velocity at the vertex of the path, then |
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Answer» `(1)/(v_(1)) + (1)/(v_(2)) = (1)/(U)` |
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| 14020. |
When are the velocity and acceleration in the same direction in SHM? |
| Answer» Solution :When a particle MOVES FRON extreme position to mean position,its VELOCITY and ACCELERATION are in the same DIRECTION. | |
| 14021. |
A bullet of mass 0.25kg is fired with velocity 302m//s into a block of wood of mass m_1=37.5kg. It gets embedded into it. The block m_1 is resting on a long block m_2 and the horizontal surface on which it is placed is smooth. The coefficient of friction between m_1 and m_2 is 0.5. Find the displacement of m_1 on m_2 and the common velocity of m_1 and m_2. Mass m_2=1.25kg. |
Answer» ` <br> <img src=) (i) COMMON VELOCITY =("Initial momentum")/("Total mass") `v_c=(0.25xx302)/(0.25+37.5+1.25)` `=1.94m//s` (II) `v_1=("Initial momentum")/(m_1+m)` `=(0.25xx302)/(37.5+0.25)=2m//s` `a_1=(F)/(m_1+m)=mug=5m//s^2` `a_2=(f)/(m_1+m+m_2)=((m_1+m)/(m_1+m+m_2))mug` `=((37.5+0.25)/(37.5+0.25+1.25))(0.5xx10)` `=4.84m//s^2` `a_r=a_1-a_r=0.16m//s^2` Common velocity is achieved when, `v_1` converts into `v_c` by a retardation `a_1`. `:. v_c=v_1-a_1t` `:. t=(v_1-v_c)/(a_1)=(2-1.94)/(5)` `=0.012s` Now, `s_r=1/2a_rt^2` `=1/2xx0.16xx(0.012)^2` `=0.011mm` |
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| 14022. |
A car is driven round a curved path of radius 18m in without the danger of skidding. The coefficient of friction between the tyres of the car and the surface of the curved path is 0.2. What is the maximum speed in kmph of the car for safe driving ? (g=10 ms^(-2)) |
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Answer» 21.6 KMPH |
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| 14023. |
A car of mass 2 metric tons is moving up a smooth inclined plane with a velocity of 72KMPH. If the power of the engine is 8KW, the angle of inclination of the inclined plane is |
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Answer» `Tan^(-1) ((1)/(49))` |
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| 14024. |
In the arrangement shown in show that tension in the string between masses m_(2) and m_(3) is T = (2 m_(1) m_(3) g)/(m_(1) + m_(2) + m_(3)) |
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Answer» SOLUTION :Let `T_(1)` be the tension in the STRING between `m_(1)` and `m_(2)` and `T` be the tension in the string between `m_(2)` and `m_(3)` If a is acceleration of the SYSTEM in the direction shown then `m_(1) a =T_(1) - m_(1) g` ….(i) `(m_(2) + m_(3)) a = (m_(2) + m_(3)) g - T_(1)` ....(ii) and `m_(3) a = m_(3) g - T` ...(iii) Adding (i) and(ii) `(m_(1)+ m_(2) +m_(3) a = (m_(2) + m_(3) - m_(1)) g` `a = (m_(2) + m_(3) - m_(1)) g// (m_(1) + m_(2) + m_(3))` PUT in (iii) `(m_(3) (m_(2) + m_(3) -m_(1)) g)/(m_(1) + m_(2) + m_(3)) =m_(3) g - T` `T = m_(3) g-(m_(3) (m_(2) + m_(3) -m_(1)) g)/(m_(1) + m_(2) + m_(3))` `T = 2 m_(1) m_(3) g//(m_(1) + m_(2) + m_(3))` . |
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| 14025. |
An ideal gas undergoes four different processes from the same initial state as shown in figure of P-V diagram. Four processes are adiabatic, isothermal, isobaric and isochoric. Out of 1,2,3 and 4 which one is adiabatic ? |
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Answer» 4 |
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| 14026. |
From a complete ring of mass M and radius R, an arc of making 30^(@) at centre is removed. What is the moment of inertia of the incomplete ring about an axis passing through the centre of the ring and bot^(er) to the plane of the ring |
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Answer» Solution :mass of in COMPLETE ring = `M-(M)/(2pi)xx (pi)/(16)=(11M/(12))` MOI of inertia of INCOMPLETE ring `I=[(11m)/(12)]R^(2)` 
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| 14027. |
In Column-I process and in Column-II first law of thermodyanmics are given . Match them appropriately : |
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Answer» |
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| 14028. |
Efficiency of a Carnot engine is 50% when temperature of outlet is 500 K. In order to increaseefficiency up to 60% keeping temperature of intake the same what is temperature of outlet |
| Answer» Answer :B | |
| 14029. |
A ball is dropped at t = 0 from a height 12 m above the ground. At the instant of release a very massive platform is at a height 4 m above the ground and moving upward with velocity 3 m//s as shown in the figure. Find : (a) the height reached by the ball after a perfectly elastic impact with the wall. the time when the ball strikes the platform second time. |
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| 14030. |
Define Torque and derive its expression. |
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Answer» Solution :(i) Torque is defined as the moment of the external APPLIED force about a point or axis of rotation. (ii) `VEC(tau) = vec(r) xx vec(F)` where `vec(r)` is the position vector of the point and the force `vec(F)` is acting on the body as shown in FIGURE.  (iii) Here, the product of `vec(r) and vec(F)`is called the vector product or cross product. The vector product of two VECTORS results in another vector that is perpendicular to both the vectors. Hence, torque `(vec(tau))` is a vector quantity. (iv) Torque has a magnitude (r F `sin THETA`) and direction perpendicular to `vec(r) and vec(F)`. Its unit in N m. `vec(tau) = (r F sin theta) hat(n)` (v) Here, `theta` is the angle between `vec(r) and vec(F) and hat(n)` is the unit vector in the direction of `(vec(tau))`. |
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| 14031. |
Match List I with List II and select the correct answer |
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Answer» pascal = unit of PRESSURE = `(F)/(A)=( [MLT^(-2)])/([L^(2)])= [M^(1)L^(-1)T^(-2)]` hertz = unit of frequency =`(1)/(T)= [M^(0)L^(0)T^(-1)]` JOULE = unit of work = FORCE x distance= `[MLT^(-2)][L]= [M^(1)L^(2)T^(-2)]` |
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| 14033. |
A cyclist starts from the centre O of acircular park of radius 1 km, reaches the adge P of the park , then cycles along the circumference , and returns to the centre along QO as shown in Fig . 4.21 . Ifthe round trip takes 10 min , what is the (a)net displacement. (b) average velocity , and (c ) average speed of the cyclist ? |
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Answer» |
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| 14034. |
Sound travels faster on a rainy day than on a dry day. Why? |
| Answer» Solution :The AMOUNT of water vapours present in the ATMOSPHERE is much higher on a rainy DAY than on a dry day. As the water vapours are lighter than dry AIR, density of wet air becomes less than of dry air. Because the speed of sound is INVERSELY proportional to the square root of the density, sound travels faster on a, rainy day than on a dry day. | |
| 14035. |
The acceleration due to gravity on the surface of the moon is 1.7 ms^(-2). What is the time period of a simple pendulum on the surface of moon if its time period on the surface earth is 3.5 s. ? (g on the surface earth 9.8 ms^(-2)) |
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Answer» SOLUTION :Here, `g_m = 1.7 MS^(-2) , g_e = 9.8 ms^(-2) , T_e = 3.5 s^(-1)` As `T_e = 2pi SQRT(L/(g_e)) and T_m = 2pi sqrt(l/g_m)` `therefore (T_m)/(T_e) =sqrt((g_e)/(g_m)) ` (or) `T_m = T_e sqrt((g_e)/(g_m)) = 3.5 sqrt((9.8)/(1.7)) =8.4 s` |
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| 14036. |
The magnitude of gravitational field at distance r_(1) and r_(2) from the centre of a uniform sphere of radius R and mass M are I_(1) and I_(2) respectively. Then |
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Answer» `(I_(1))/(I_(2))(r_(1))/(r_(2))` if `r_(1) lt R and r_(2)_ lt R` |
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| 14037. |
The pendulum of certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours? |
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Answer» Now no of OSCILLATION in 1 day `=(24xx3600)/2` =43200 But in each oscillation it is slower by `(2.04-2.00)=00.04sec` So in ONE day it is slower by `=43200xx(0.04)` `=28.8min` so, the clock runs 28.8 minutes slowere in one day. |
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| 14038. |
Three samples of the same gas A,B and C (gamma=3//2) have initially equal volume. Now the volume of each sample is doubled. The process is adiabatic for A, isobaric for B and isothermal for C. If the final pressures are equal for all three samples, find the ratio of their initial pressures. |
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Answer» Solution :LETE the INITIAL pressure of the THREE samples be `P_(A), P_(B) and P_(C)` then `P_(A)(V)^(3/2)=(2V)^(3/2)P` `P_(B)=P ""P_(C)(V)=P(2V)` ` :. P_(A):P_(B):P_(C) = (2)^(3//2): 1:2 = 2sqrt2:1:2` |
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| 14039. |
An object falling through a fluid is observed to have an acceleration given by a=g-bv where g is the gravitational acceleration and b is a constant. After a long time it is observed to fall with a constant velocity. What would be the value of this constant velocity? |
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Answer» `g/b` |
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| 14040. |
Planck's constant has the same dimensions as |
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Answer» Energy |
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| 14041. |
The resultant of two like parallel forces is 14 kgwt and acts at distance of 30 cm and 40 cm from them. The forces are |
| Answer» Answer :A | |
| 14042. |
The increase of length of a wire under a load is equal to its initial length. What is the stress in the wire equal to? |
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Answer» |
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| 14043. |
The breaking stress for a wire of unit cross - sectional is called its |
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Answer» yeild POINTS |
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| 14044. |
The simplest example of SHM is the oscillation of a simple pendulum In a simple pendulum made of a metallic wire, what will happen to period when temperature increases ? give reason. |
| Answer» SOLUTION :As TEMPERATURE INCREASES LENGTH increases so PERIOD increases. | |
| 14045. |
Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil ? |
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Answer» |
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| 14046. |
Find the temperature distribution in a substance placed between two parallel plates kept at temperatures T_(1) and T_(2). The plate separation is l. The heat conductivity coefficient lamda varies with temperature as 1//sqrt(T). |
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| 14047. |
Points A and B are market on the ground at a distance 1 m from each other . C is marked as midpoint between A and B . O is point on the same line at a distance 5. m from point C. A particle is projected with a speed 10 m//s from point O. What should ne the angle of projection so that particfle strikes somewhere between A and B ? |
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Answer» Solution :Rangeof the projectile can be written as follow: `R=(u^(2) sin 2 thetqa)/(g) =(10 sin 2theta)/(10)` `R=1- sin 2 theta` Range of the projectile can be plotted against `theta` as shown in the figure .  For the particle to land between A and B , its range must be between 5 m and 6m Let `alpha ` be the angle of projection corresponding to R=5m .`R=10 sinsin 2 aplha =5 rArr sin 2 alpha =1//2 rArr 2alpha =30^(@) rArr alpha 15^(@)` Range will be samefor `(90-alpha)=75^(@)` Similarly we can CALCULATE angleof projection `beta` for R=6 m . `R=10 sin 2 beta =6 rArr sin 2 beta =3//5 rArr 2 beta =37^(@) rArr beta =18.5^(@)` Rnage will be same for `(90-beta)=71.5^(@)` We can understand that range is between 5m and6m when angle of projection is between `15^(@)`and `18.5^(@)` and alsowhen angle of projection is between `71.5^(@)` and `75^(@)` . From the graph we can understand thatrange of projectile is GREATER than 6 m when angle of projectionis between `18.5^(@)` and `71.5^(@)` andthisinterval is `(71.5-18.5)=53^(@)` . Hence we can also SAY thatminimum angle of projection is `15^(@)` and maximumis `75^(@)` which is not allowed for correct landing of particle. |
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| 14048. |
Ball are dropped from the roop fo towar at fived interval if tiem . At the moment when9th ball reaches the groun the nth ball is (3//4) the heith of the tower Wgat the vale ofn ? G= 10 m//s^2. |
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Answer» SOLUTION :Let (t) be the time INTERVAL between TWO successive balls while falling . When `9th ` ball reaches the ground, the Ist ball is just to be dropped, os the time taken by the9th ball to reach the groun ` =(9 -1) t= 8 t` . `:. S= 1/2 xx g xx g xx (8t)^2 = 1/2 10 xx 64t^2 = 320 t^2 ` ...(i)` Time taken by nth ball to fall `= (n-1) t` DISTANCE travelled ` = s- 3 S/4` :. ` S/4 = 1/2 xx [(n-1)^2 t]^2 = 5 (n-a)t` or S= 20 (n-1)^2 t^2` ...(II) From (i) and (ii) ` 320 t^2 = 20 (n-1)^2t^2` On solving, ` n=5`. |
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| 14049. |
A simple harmonic motion of a particle is represented by an equation x= 5 sin(4t -(pi)/(6)), where x is its displacement. If its displacement is 3 unit then find its velocity. |
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Answer» `(2pi)/(3)` `v= omega SQRT(A^(2)-y^(2))= 4 sqrt(25-9)= 16` unit. |
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| 14050. |
When 100 J of work is done on a fly wheel, its angular velocity is increased from 60 rpm to 180 rpm. What is the moment of inertia of the wheel ? |
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Answer» SOLUTION :Kinetic energy `E=(1)/(2)I(omega_(2)^(2)-omega_(1)^(2))` `100=(1)/(2)I(4PI^(2).9-4pi^(2).1^(2))` `I=0.633kgm^(2)` |
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