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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.
| 12601. |
When a source moves at a speed greater than that of sound, will Doppler effect hold ? |
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Answer» |
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| 12602. |
A person in lift which ascents up with acceleration 10 ms^(-2) drops a stone from a height 10 m. The time of decent is [g=10ms^(-2)] |
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Answer» 1s |
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| 12603. |
In pure rolling motion f a ring (a) it rotates about instantaneous point of contact of ring and ground (b) Its centre of mass moves in translatory motion only (c)its centre of mass will have translatory as well as rotatory motion |
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Answer» only a is CORRECT |
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| 12604. |
A bullet of mass 10 gm moving with a horizontal velocity 100m//s passes through a wooden block of mass 100 gm. The block is resting on a smooth horizontal floor. After passing through the block the velocity of the bullet is 10m//s. the velocity of the emerging bullet with respect to the block is |
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Answer» `10 m//s` |
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| 12605. |
A particle executes shm of period.8 s. After what time of its passing through the mean position the energy will be half kinetic and half potential. |
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| 12606. |
At ordinary temperatures, the molecules of an ideal gas have only translational and rotational kinetic energies. At high temperatures they may also have vibrational energy. As a result of this at high temperature: (MA toMonoatomic gas & DA toDiatomic gas) |
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Answer» `C_v = (3R)/(2)` for a MA gas |
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| 12607. |
A bullet is fired from a gun. The force on the bullet is given by F = 600 - 2 xx 10^(5) t where F is in newton and't' is in second. The force on the bullet becomes zero as soon as it leaves the barrel. The average impulse imparted to the bullet is |
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Answer» 9Ns |
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| 12608. |
Which is the greatest force among the three force vecF_1 , vecF_2 , vecF_3 as shown below? |
| Answer» Solution :FORCE is a VECTOR and magnitude of the vector is REPRESENTED by the length of the vector. Here `vecF_1` has greater length COMPARED to other two. So `vecF_1` is largest of the THREE. | |
| 12609. |
The equation of a trevelling wave In a uniform string of mass per unit length unit length mu is given as y=Asin(omega-kx).Find the total energy transferred through the origin in time interval from t = 0 to t=pi//12omega.(You can use the formula in instantaneous power if you know) |
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| 12610. |
A steam engine boiler is maintained at 250^(@)C and water is converted into steam. This steam is used to do work and heat is ejected to the surrounding air at temperature 300K. Calculate the maximum efficiency it can have? |
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Answer» Solution :The steam engine is not a Carnot engine, because all the process involved in the steam engine are not perfectly reversible. But we can calculate the maximum POSSIBLE efficiency of the steam engine by considering it as a Carnot engine. `eta=1-(T_(L))/(T_(H))=1-(300K)/(523K)=0.43` The steam engine can have maximum possible `43%` of efficiency, IMPLYING this steam engine can convert `43%` of input heat into useful work and remaining `57%` is EJECTED as heat. In practice the efficiency is even less than `43%.` |
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| 12611. |
A particle at the end of the spring executes S.H.M with a period t_1, while the corresponding period for another spring is t_2. If the period of oscillation will two springs in series is 'T', then (same particle connected in all the cases) |
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Answer» `T = T-1 + t_2` |
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| 12612. |
Temperature gradient is negative if a cube block is heated at cold end and heat transfers to hot end. |
| Answer» SOLUTION :Here, TEMPERATURE GRADIENT is POSITIVE. | |
| 12613. |
A disc of mass 0.5 kg is kept floating horizontally in mid air by firing bullets of mass 5 g each, vertically at it at the rate of 10 per second. If the bullets drop dead. The speed of the bullet striking the disc is (g=10 ms^(-2)) |
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Answer» `100 MS^(-1)` |
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| 12614. |
Find the radio of the orbital speed to two satellites of the earth if the satellites are at height 6400km and 19200km. (Radius of the earth = 6400km.) |
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Answer» Solution :Orbital speed , `v_(0)=sqrt((GM)/(R+h))` The RATIO of the orbital speeds of two SATELLITES around the earth is `(V_(01))/(V_(02))=sqrt((R+h_(2))/(R+h_1))` Where `h_1 and h_2` are the heights of the satellites In this PROBLEM `h_1 = 6400km,h_(2) = 19200 km` R = 6400 km. `(V_(01))/(V_02)=sqrt((6400+19200)/(6400+6400))=sqrt((25600)/(12800))=sqrt2/1` |
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| 12615. |
A body rolls down a stair case of 5 steps. Each step has height 0.1 m and width 0.1 m. With what velocity will the body reach the bottom? |
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Answer» 6.25 m |
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| 12616. |
Let I_(1) and _(2), be the moments of inertia of two bodies of identical geometrical shape, the first made of aluminium and the second of iron. |
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Answer» `I_(1) LT I_(2)` |
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| 12617. |
How is physics learning really exiting ? |
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Answer» Solution :(i) The basic concepts and LAWS of Physics can explain diverse physical phenomena. (ii) The most interesting part is the designing of useful devices based on the physical laws. Eg: (a) use of robotics, (b) journey to Moon and to nearby planets with controls from the GROUND, (c) technological advances in health sciences etc. (iii) Carrying out CHALLENGING new experiments to unfold the secrets of nature and in verifying or falsifying the existing THEORIES. (iv) Physics helped human beings in many ways LIKE the usage of laser in bloodless surgeries. |
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| 12618. |
A particle executes SHM of period 22 second and amplitude 10 cm. Find the distance it travels in 3.5 second starting from the point of zero displacement. |
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Answer» Solution :Amplitude of the particle = 10cm =0.01 m, Time period t= 22s and `phi= 0` The angular velocity , `omega= (2pi)/(T)= (2xx 22)/(7xx 22)= (2)/(7)rad//s` The equation for displacement of the particle is `x= ASIN(omegat+phi)= 0.10sin((2)/(7))t` When `t= 3.5 s= (7)/(2)s` the distance it TRAVEL becomes EQUAL to the displacement because the given time is less than `T//4` `:. x= 0.10sin((2)/(7) xx (7)/(2))= 0.10 sin(1)` `= 0.10 sin(57^(@)17.)( :. 1 "radian"= 57^(@)17.)= 0.10 xx 0.8414= 0.08414m` |
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| 12619. |
Ifthe sizeofthe nucleusis inthe range of10 ^ (-15 )mto10 ^( -14 )m andisscaledup tothetip ofasharppin, whatisroughlythesizeofanatom ? Assumethe tip ofthe pinto beintherange10 ^(-5)mto10 ^( -4 )m |
| Answer» SOLUTION :Sizeofthe ATOMIS` 10 ^(-11) `m | |
| 12620. |
Show that internal forces can change the kinetic energybut not linear momentum . |
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Answer» Solution :`vecF_("ext") = M veca_(CM) implies vecF_("ext") = (M d vecv_(CM))/(dt)` `vecF_("ext") = (d)/(dt) (M vecv_(CM))`  Only external forces can change the MOMENTUM of a SYSTEM . Internal forces can gives rise to kinetic energy . A compressed SPRING attached to two blocks , converts its potential energy into kinetic energy . This K.E. is due to internal forces . |
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| 12621. |
When a car of mass 1200 kg is moving with a velocity of 15 ms^(-1) on a rough horizontal road, its engine is switched off. How far does the car travel before it comes to rest if the coefficient of kinetic friction between the road and tyres of the car is 0.5 ? (g=10 ms^(-2)) |
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Answer» 21.6 m |
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| 12622. |
Heat is supplied to a certain homogeneous sample of matter at a uniform rate.Its temperature is plotted against time as shown in fig. Which of the following conclusion can be drawn? |
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Answer» Its SPECIFIC HEAT capacity is greater in SOLID state than in liquid state |
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| 12623. |
An ideal gas mixture filled inside a balloon expands according to the relation PV^(2/3) = constant. What is the temperature inside the balloon. |
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Answer» Solution :`PV^(2//3) = "CONSTANT" rArr (PV^(2//3))/(PV)=(constant)/(RT) or 1/V^(1/3)="constant"/(RT) rArr V prop T^(3)` TEMPERATURE increases with INCREASE in volume. |
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| 12624. |
If a stone is released from a tower, then its total energy during its fall. |
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Answer» INCREASES |
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| 12625. |
A bimetallic strip is formed out of two identical strips. One of copper and the other of brass. The coefficients of linear expansion of the two metals are alpha_(C ) and alpha_(B). On heating. the temperature of the strip goes up by DeltaT and the strip bends to form a are of radius of curvature R. Then R is: |
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Answer» Proportional to `DELTA T` |
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| 12626. |
A constant volume gas thermometer shows pressure readings of 50 cm and 90 cm of mercury at 0^(0)Cand 100^(0)C respectively. What is the , . temperature on gas scale when the pressure reading is 60 cm of mercury ? |
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Answer» Solution :GIVEN that `P_(0) = 50` cm of Hg `p_(100) = 90 ` cm of He P = 60CM of Hg t= ? `t_(C) = (P - P_(0))/(P_(100) - P_(0)) XX 100^(0 ) C ` so `t_(C) = (60 - 50)/(90 - 50) xx 100^(0) C = 26^(0) C ` |
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| 12627. |
As shown in adjacent figure if a load of mass (m) is attached at lower end of lower wire. Then find the displacementsof the points B, C, D are |
Answer» Solution : As shown in FIGURE elongation of first wire `e_(1)=((mg)l_(1))/(Ay_(1))` elongation of 2nd wire `e_(2)=((mg)l_(2))/(Ay_(2))` elongation of 3rd wire `e_(3)=((mg)l_(3))/(Ay_(3))` DISPLACEMENT of B is `e_(1)` displacement of C is `e_(1)+e_(2)` displacement of D is `e_(1)+e_(2)+e_(3)` |
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| 12628. |
A small ball strikes a stationary uniform rod, which is free to rotate, in gravity free space. The ball does not stick to the rod The rod will rotate about |
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Answer» its centre of mass |
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| 12629. |
An object is weighed at the North Pole by a beam balance and a spring balance, giving readings of W_B and W_S respectively. It is again weighed in the same manner at the equator, giving reading of W._B and W._S respectively. Assume that the acceleration due to gravity is the same everywhere and that the balances are quite sensitive (a)W_B=W_S , (b)W._B=W._S (c)W_B=W._B , (d)W._S lt W_S |
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Answer» only a & B are true |
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| 12630. |
A block is placed on a rough inclined plane of inclination theta = 30^(@). If the force to drag it along the plane is to be smaller than to lift it. The coefficient of friction mu should be less than |
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Answer» Solution :`MG(sin theta + MU cos theta)lt mg` `sin 30^(@)+mu cos 30^(@)lt 1` `(1)/(2)+mu(sqrt(3))/(2)lt 1` `RARR mu (sqrt(3))/(2)lt 1//2 rArr mu lt (1)/(sqrt(3))` |
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| 12631. |
A student had a breakfast of 200 food calories. He thinks of burning this energy by drawing water from the well and watering the trees in his school. Depth of the well is about 25 m. The pot can hold 25 L of water and each tree requires one pot of water. How many trees can he water? (Neglect the mass of the pot and the energy spent by walking. ("Take g" = 10 ms^(-2)) |
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Answer» Solution :To draw 25 L of water from the well, the student has to do WORK against gravity by burning his energy. Mass of the water `=25L=25kg(1L=1kg)` The work required to draw 25 kg of water = gravitational potential energy GAINED by water. `W=mgh=25xx10xx25=6250J` The total energy gained from the food = 200 food cal = 200 kcal `=200xx10^(3)xx4.186J=8.37xx10^(5)J` If we assume that by using this energy the student can drawn .n. pots of water from the well the totalenergy spent by him `=8.37xx10^(5)J=nmgh` `n=(8.37xx10^(5)J)/(6250J)~~134` This n is also equal to the number of trees that he can water. Is it possible to draw 134 pots of water from the well just by having breakfast? No. Actually the human body does not convert entire food energy into work. It is only approximately `20%` EFFICIENT. It implies that only `20%` of 200 food calories is used to draw water from the well. So `20 %` of the 134 is only 26 pots of water. It is QUITE meaningful. So he can water only 26 trees. |
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| 12632. |
The vein of healthy person is horizontal and its length is 0.04m and has internal radius Imm. At 37^(@)C, the average velocity of blood through the vein is 4cm s^(-1) Find. (i) the rate of flow of the blood (ii) the velocity at the mid point of the verin and (iii) the decreases in pressure at th other end of the vein. Take eta_("blood")=4xx10^(-3)Pa-sec |
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Answer» `1.25xx10^(-7), m^(3)s^(-1), 0.08 m s^(-1), 50.9 PA` |
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| 12633. |
Spring is pulled down by 2 cm. What is amplitude of motion? |
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Answer» 0 cm |
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| 12634. |
The static friction comes into play at the moment the force is applied. Write the relation between static friction and normal reaction. |
| Answer» SOLUTION :`f^max_s = mu_s N` | |
| 12635. |
The periodic time of a particle doing simple harmonic motion is 4 second. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is |
| Answer» ANSWER :D | |
| 12636. |
What is the acceleration due to gravity on the surface of a planet that has twice the mass and radius of the earth? |
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Answer» SOLUTION :`IMPLIES` For PLANET `g_p=(GM_p)/(R_P^2)` `g_p=(G(2M_e))/(4R_e^2)` `=1/2((GM_e)/(R_e^2))` `g_p=g/2` |
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| 12637. |
The piston and nozzle of a syringe kept horizontal have diameters 5 mm and 1 mm. the piston is pushed with constant velocity of 0.2ms^(-2) Find the horizontal distance travelled by water jet before touchig water[g=10m//s^(2)]. Height of syringe from ground is 1 m. |
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Answer» Solution :`d_(1)=5mmthereforer_(1)=2.5mm=2.5xx10^(-3)m` `d_(2)=1mmthereforer_(2)=0.5mm=0.5xx10^(-3)m` If velocity of water coming out of a SYRINGE is `v_(2)` , according to CONTINUITY equation , `A_(1)v_(1)=A_(2)v_(2)` `thereforev_(2)=((A_(1))/(A_(2)))v_(1)` `=((pir_(1)^(2))/(pir_(2)^(2)))v_(1)=((r_(1))/(r_(2)))^(2)v_(1)` `=((2.5)/(0.5))^(2)xx0.2(becausev_(1)=0.2m//s))` `thereforev_(2)=5m//s`. Time taken by water jet when it touches the ground using. `d=(1)/(2)g t^(2)` `1=(1)/(2)xx10t^(2)(d=1.0m)` `thereforet=(1)/(sqrt(5))s` `therefore`Horizontal distance covered by water jet in time t is , `x=v_(2)t=5xx(1)/(sqrt(5))` `thereforex=sqrt(5)m`. `thereforex=2.23m`. |
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| 12638. |
A bob is suspended from an ideal string of length 'l'. Now it is pulled to side through 60^(@) to vertical and whirled along a horizontal circle. Then its period of revolution is |
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Answer» `PI SQRT(l//g)` |
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| 12639. |
Two particles of masses 10 kg and 30 kg are lying on a stright line. The 10 kg mass is shifted towards the 30 kg mass by a distance of 2cm. By what distance should the 30 kg mass be shifted so that the position of their centre of mass does not change |
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Answer» 2/3 cm towards 10 kg |
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| 12640. |
The earth rotates from west to east. A wind mass begains moving due north from the equator, along the earth's surface Neglect all effects other than the rotation of the earth . The wind mass will |
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Answer» always move due north |
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| 12641. |
One end of string of length 1m is tied to a body of mass 0.5 kg. It is whirled in a vertical circle as shown in figure. If the angular frequency of the body is 4rad/sec |
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| 12642. |
A mass m is released from the top of a vertical circular track of radius r with a horizontal speed v_(0). Calculate the angle theta with respect o the vertical where it leaves contact with the track. |
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Answer» Solution :The forces acting on the body are its weight mg and reaction N as shown in Fig. So for circular motion of the body at any posiotion `theta`  `(mv^(2))/(r )= mg cos theta - N` or `N = mg cos theta - (mv^(2))/(r )` The body will leave contact where N = 0 i.e., `mg cos theta = (mv^(2))/(r )=0` i.e., `cos theta =(v^(2))/(RG)` .....(2) Now applying CONSERVATION of mechanical energy between H and P, we get `(1)/(2) mv^(2)=(1)/(2)mv_(0)^(2)+mgr (1-cos theta)` where .v. is the velocity at .P. [as `y=r (1-cos theta)]` or `v^(2)=v_(0)^(2)+2gr (1-cos theta)` ......(3) substituting the value of `v^(2)` in Equation (2) `cos theta = (v_(0)^(2))/(rg)+2(1-cos theta)` i.e., `cos theta = [(v_(0)^(2))/(3rg)+(2)/(3)] or theta = cos^(-1) [(v_(0)^(2))/(3rg)+(2)/(3)]` |
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| 12643. |
Find the CG of :(a) a cone of height h from the base,(b) a hemisphere of radius r from its flat surface. |
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| 12644. |
Two metal plates A and B made of same material are placed on a table as shown in the figure. If the plates are heated uniformly, will the gap indicated by x and y in the figure increase or decrease? |
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| 12645. |
Two bodies of masses 2kg and 8kg are at rest. If same force acts on them, the ratio of distances travelled by them before they attain same kinetic energy is |
| Answer» ANSWER :A | |
| 12646. |
Calculate the rate of increment of the thickness of ice layer on a lake when thickness of ice is 10 cm and the air temperature is -5^(@) C. If thermal conductivity of ice is 0.008 cal "cm"^(-1) "s"^(-1)°"C"^(-1), density of ice is 0.91 xx 10^(3) "kg m"^(-3) and latent heat is 79.8 cal "gm"^(-1). How long will it take the layer to become 10.1 cm ? |
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Answer» Solution :The time taken for the layer of ice to increase its thickness from `x_1` to `x_2` is given by `int_(0)^(t) DT = (rho L)/( K THETA ) int_(x_1)^(x_2) X" "dx` `t = ( rho L)/( 2 K theta ) [ x_(2)^(2) - x_(1)^(2) ]` Given `theta = 5°C, rho = 0.91 "GM/c.c." L= 79.8 "cal gm"^(-1)` `x_1 = 10 "cm", x_2 = 10.1 "cm", K = 0.008 "cal cm"^(-1) s^(-1)°"C"^(-1)` Rate of increment of thickness`= (dx)/(dt) = (K ( theta_2 - theta_1) )/(x rho L) = (0.0008 xx 5)/( 10 xx 0.91 xx 79.8) = 5.5 xx 10^(-5) "cm/s"` `t = (0.91 xx 79.8)/( 2 xx 0.008 xx 5 ) [ 10.1^(2) - 10^(2) ] = 1824.527` S |
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| 12647. |
Centre of mass of two particles with masses 2 kg and 1 kg located at (1,0,1) and (2,2,0) has the co-ordinates of |
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Answer» `(4)/(3),(2)/(3),(2)/(3)` |
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| 12648. |
A uniform solid sphere of mass 0.5kg rolls on a horizontal surface without slipping. If the velocity of centre of mass is 4 cm s^(-1), the work done in stopping sphere is |
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Answer» `8 XX 10^(-4) J` |
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| 12649. |
The magnitude of gravitational field at distances r_(1) and r_(2) from the center of a uniform sphere of radius R and mass M are respectively I_(1) and I_(2). The ratio I_(1)//I_(2) is given in the column II. Make correct matching {:("Column-I",,"Column-II"),((A) r_(1) gt R and r_(2) gt R,,(P) (r_(1)r_(2)^(2))/(R_(3))),((B) r_(1) lt R and r_(2) gt R,,(Q) (R_(3))/(r_(2)r_(1)^(2))),((C) r_(1) gt R and r_(2) lt R,,(R) (r_(1))/(r_(2))),((D) r_(1) lt R and r_(2) gt R,,(S)(r_(2)^(2))/(r_(1)^(2))):} |
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