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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. |
A bullet of mass 0.005 kg moving with a speed of 200 ms^(-1) enters a heavy wooden block and is stopped after a distance of 50 cm. What is the average force exerted by the block on the bullet ? |
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Answer» `-200 N` |
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| 2. |
A copper calorimeter containing water is heated by means of an electric heater for a certain time. Next, the water is transferred to a container having negligible heat capacity, and the heater is operated again for the same time. The rise in temperature of water in the latter case is observed to be 1.2 times that in the former, what is the ratio of the water equivalent of the copper calorimeter to the mass of water ? |
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Answer» |
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| 3. |
The kinetic theory of gases breaks down most at ............... . |
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Answer» LOW PRESSURE and HIGH temperature |
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| 4. |
A wave propagates on a string in the positive x-direction at a velocity u. The shape of the string at t=t_0is given by g(x, t_0)= A sin(x /a) . Write the wave equation for a general time t. |
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Answer» For a WAVE travelling in the positive x-direction the general EQUATIONN is given by `y=Asin(x/a-t/T)` Putting t=-t and comparing with equation 1 we get `RARR G(x,0)=Asin{(x/a)+(t_0/t)}` `rarr g(x,t)=Asin[{(x/a)+t_0/T}-(t/T)]` As T=a/v` a=wave LENGTH v=speed of the wave `rarr y=A sin[(x/y)+t_0/((a/v))-t/((a/v))` `rarr y=Asin(x+v(t_0-t))/a` |
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| 5. |
Same ideal gas is kept in two containers A and B fitted with frictionless pistons. Volume and temperature of the gas in both containers are the same. m_A and m_B are the masses of the gas in A and B respectively.Volume of the gases in the two containers are changed to 2V keeping their temperature constant. Corresponding changes in pressure in A and B are Deltap and 1.5Deltap. Find the ratio of the masses of the gas kept in A and in B. |
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Answer» Solution :Volume of the gases in both the CONTAINERS become double. So,according to Boyle.s law CORRESPONDING pressures of the gases become HALF of their initial pressure. `therefore` Initail pressure in A and B are `p_A=2 Deltap` and `2 times 1.5 Deltap=3 Deltap` `thereforep_A/p_B=2/3or,(n_A(RT)/V)/(n_B(RT)/V)=2/3` or,`(m_A(RT)/(MV))/(m_B(RT)/(MV))=2/3or,m_A/m_B=2/3.` |
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| 6. |
A cylinder of mass M and radius R starts falling under gravity at t = 0 as shown in figure. If the mass of chord is negligible, the tension in each string is: |
| Answer» ANSWER :A | |
| 7. |
Three of the quantities defined below has the same dimensional formula. Identify the (i) sqrt("Energy/mass") (ii) sqrt("pressure/density") (iii) sqrt("force/linear density") (iv) sqrt("Angular frequncy/radius") |
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Answer» II only |
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| 8. |
A body is projected vertically up with a velocity of 58.8 m/s. After 3 s if the acceleration due to gravity of earth disappears. The distance travelled and the velocity of the body at the end of next 5 sec is |
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Answer» 147 m, 29.4 m/s |
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| 9. |
Find the moment of inertia of a thin uniform rod about an axis perpendicular to its length and passing through a point which is at a distance of (l)/(3) from one end. Also find radius of gyration about that axis. |
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Answer» Solution :(i) Let P be the point at a distance `(L)/(3)` from ONE end. It is at a distance of `((l)/(2)-(l)/(3))=(l)/(6)` from the CENTRE as shown in the figure. By parallel axes theorem `I=I_(c )+MR^(2)` `=(Ml^(2))/(12)+M((l)/(6))^(2)`  (ii) The radius of gyration `K=SQRT((I)/(M))=sqrt((Ml^(2))/(9M))=(l)/(3)` |
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| 10. |
The working of an atomizer depends upon |
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Answer» BERNOULLI's theorem |
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| 11. |
A block of mass m is kept on the ground. A vertically upward for F = kt (K is a positive constant) where 't' is in seconds , starts to act on it at t = 0 |
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Answer» the block looses contact with ground at t = mgK |
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| 12. |
A satellite is revolving round the earth in a circularorbit with a velocity of 8 km/s . At a height where acceleration due to gravity is 8m//s^2 . How high is the satellite from the earth ? (Take R=6000 km) |
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Answer» Solution :As centripetal ACCELERATION equals to acceleration DUE to gravity at that HEIGHT , then `therefore a=V^2/r=g_h rArr V^2/r = 8 = (64xx10^6)/(R+h)=8` `rArr R+h=8xx10^6` `h=8xx10^6 - 6xx10^6 =2xx10^6` m = 2000 km |
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| 13. |
Radius of curvature of first surface of double convex lens is three times that of the other. If focal length of the lens is 30cm and refractive index of the lens is 3//2. Then radius of curvature of the first surface is |
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Answer» 20cm |
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| 14. |
The origin of friction is |
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Answer» ELECTROSTATIC INTERACTION |
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| 15. |
A 1 mlong rod having a constant cross sectional area is made of four materials. The first 0.2 m is made of iron, the next 0.3 m of lead, the following 0.2m of aluminium and the remaining part is made of copper. Find the centre of mass of the rod. The densities of iron, lead, aluminium and copper are 7.9 xx 10^3kg//m^3, 11.4 xx 10^3 kg//m^3, 2.7 xx 10^3kg//m^3 and 8.9 xx 10^3 kg//m^3 respectively. |
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Answer» Solution :According to the definition of density we can write, `d=m/v` `rArr m = v xx d` `rArr m=A xx l xx d [therefore V = A xx l]` In case of uniform rod its area of cross section is constant. Hence `m PROP l xx d [ therefore " A is constant "]` MASS of IRON part, `m_(1) = 0.2 xx 7.9 xx 10^(3)=1.58 xx 10^(3) kg` Mass of lead part, `m_(2) prop 0.3 xx 11.4 xx 10^(3) = 3.42 xx 10^(3)` kg Mass of ALUMINIUM part, `m_(3) prop 0.2 xx 2.7 xx 10^(3) = 0.54 xx 10^(3) kg` Mass of copper part, `m_(4) prop 0.3 xx 8.9 xx 10^(3) = 2.67 xx 10^(3)` kg CM of iron part from end "A" of the rod, `x_(1)=0.1`m CM of lead part from end "A" of the rod, `x_(2)= 0.35 m` CM of aluminium part from end "A" of the rod, `x_(3) = 0.6 m` CM of copper part from end "A" of the rod, `x_4= 0.85m` `therefore`Centre of mass of the rod, `X_( c)=(m_(1)x_(1) + m_(2)x_(2) + m_(3)x_(3) + m_(4)x_(4))/(m_(1) + m_(2) + m_(3) + m_(4))` `rArr X_(cm) = 0.481 m`from the end "A" of the rod. 
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| 16. |
20g of water is enclosed in a thermally insulated cylinder at a temperature of 0^(@)C under a weightless piston whose area is s=500cm^(2). The outside pressure is equal to standard atmospheric pressure. To what height will the piston rise when water absorbs Q=20kJ of heat? Sp. heat of water =4200J//kg//K, sp. latent heat of water =2250kJ//kg and boiling point of water =373K |
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Answer» |
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| 17. |
Select the correct pairs of factors on which does the velocity of sound in any gas depend. |
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Answer» DENSITY and ELASTICITY of gas |
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| 18. |
A block of mass 'm' is resting on the floor of a lift. The coefficient of friction between the block and the floor is mu. When the lift is falling freely, the limiting frictional force between block and surface is |
| Answer» ANSWER :D | |
| 19. |
A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.5, the maximum percentage of the length of the chain that can hang over one edge of the table is |
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Answer» `20%` |
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| 20. |
Suppose refractive index mu is given as mu= A + (B)/(gamma^(2))where A and B are constants and gamma is wavelength, then dimensions of B are same as that of |
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Answer» wavelength hence `mu` is dimensionless, THUS each term on the RHS of given equation should be dimensionless. `:. (B)/(lambda^(2))` is dimensionless, i.e, B should have dimensions of `lambda^(2)` i.e, `L^(2)` area. |
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| 21. |
An object executing SHM has an acceleration of 0.2m//s^(2) at a distance of 0.05m from the mean position. Calculate the period of SHM. |
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Answer» 5.764 sec |
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| 22. |
If the distance between the centres of the atoms of potassium and bromine in KBr (potassium bromide) molecule is 0.282 xx 10^(-9)m, find the centre of mass of this two particle system from potassium (mass of bromine = 80 u, and of potassium = 39 u). |
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| 23. |
(A) : On a curved path average speed of a particle can never be equal to average velocity. ( R) : Average speed is total distance travelled divided by total time. Whereas average velocity is, final velocity plus initial velocity divided by two. |
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Answer» Both (A) and ( R) are ture and ( R) is the correct EXPLANATION of (A) |
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| 24. |
Two bodies of masses .m,. and .m,. are moving with velocity v _(1) and .v_(2). respectively in the same direction. The total momentum of the system in the frame of reference attached to the centre of mass is (v is relative velocity between the masses) mm,v |
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Answer» `(m_(1)m_(2)v)/(m_(1)-m_(2))` |
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| 25. |
IF the coefficient of volume expansion of a solid, a liquid and a gas are gamma_s,gamma_l and gamma_g respectively then |
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Answer» for DIFFERENT solids ,LIQUIDS and gases the values of `gamma_s,gamma_l` and `gamma_g` are different |
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| 26. |
Give example of a motion wherex gt 0 , v lt 0, a gt 0at a particular instant. |
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Answer» Solution :Let the motion is represented by `x (t) = 4 + 5e ^(- gammat) ""…(1)` Let `A gt B and gamma gt 0` Now velocity `x (t) = (dx)/(dt) =- 5 (7) E ^(-7t)` Acceleration `a (t) = (dx)/(dt) = 5 (7) ^(2) e^(-7t)` Suppose we are considering any instnat t, then from Eq. (i) we can SAY that `x (t) gt 0, v (t) lt 0 and a gt 0.` |
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| 27. |
A scientist says that the efficiency of its his heat engine which operates at source temperature 127^(@)C and sink temperature 27^(@)C is 26%, then |
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Answer» It is impossible |
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| 28. |
Supernova refer to the explosion of a massive star. The material in the central case of such a star continues to collapse under its own gravitational pull. If mass of the core is less is than 1.4 times the mass of sun, its collapse finally results in a white dwarf star. However, if the core has a mass greater than this, it could end up soon as a neutron. star and if its mass is more thanabout three solar masses, the collapse may still continue till the star becomes a very small object with an extremely high value of density called a 'Black hole'. Escape speed for a black hole is very large. The figure shows a block hole of radius R and another concentric sphere of radius R_(S), called the 'Schwarzschild Radius'. It is the critical radius at which escape speed equals the speed of light c. Nothing even the light, can escape from within the sphere of Radius R_(S). So light from a black hole can not escape and hence the terminology 'black hole'. There has been astronomical evidence of a small and massive object at the centre of our galaxy the 'Milky way'. Suppose that there is a particle at a distance about 6 light years. that orbits this massive object with an orbital speed of about 2 xx 10^(5) m//s. Use the given data whereever necessary and answer the questions that hollow. G = 6.67 xx 10^(-11) N - m^(2)//kg^(2), Solar mass M = 2 xx 10^(30) kg, C = 3 xx 10^(8) m//s, 1 light year = 9.5 xx 10^(15) m. Mass (in kg) of the massive object at the centre of the milky galaxy is of the order : |
| Answer» Answer :B | |
| 29. |
Supernova refer to the explosion of a massive star. The material in the central case of such a star continues to collapse under its own gravitational pull. If mass of the core is less is than 1.4 times the mass of sun, its collapse finally results in a white dwarf star. However, if the core has a mass greater than this, it could end up soon as a neutron. star and if its mass is more thanabout three solar masses, the collapse may still continue till the star becomes a very small object with an extremely high value of density called a 'Black hole'. Escape speed for a black hole is very large. The figure shows a block hole of radius R and another concentric sphere of radius R_(S), called the 'Schwarzschild Radius'. It is the critical radius at which escape speed equals the speed of light c. Nothing even the light, can escape from within the sphere of Radius R_(S). So light from a black hole can not escape and hence the terminology 'black hole'. There has been astronomical evidence of a small and massive object at the centre of our galaxy the 'Milky way'. Suppose that there is a particle at a distance about 6 light years. that orbits this massive object with an orbital speed of about 2 xx 10^(5) m//s. Use the given data whereever necessary and answer the questions that hollow. G = 6.67 xx 10^(-11) N - m^(2)//kg^(2), Solar mass M = 2 xx 10^(30) kg, C = 3 xx 10^(8) m//s, 1 light year = 9.5 xx 10^(15) m. If mass of earth M_(E) ~~ 6 xx 10^(24) kg and its radius R_(E) = 6400 km, to what fraction of its presents radius does the earth need to be compressed in order to become a black hole ? (Give only the order of your answer) |
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Answer» `10^(-4)` |
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| 30. |
Supernova refer to the explosion of a massive star. The material in the central case of such a star continues to collapse under its own gravitational pull. If mass of the core is less is than 1.4 times the mass of sun, its collapse finally results in a white dwarf star. However, if the core has a mass greater than this, it could end up soon as a neutron. star and if its mass is more thanabout three solar masses, the collapse may still continue till the star becomes a very small object with an extremely high value of density called a 'Black hole'. Escape speed for a black hole is very large. The figure shows a block hole of radius R and another concentric sphere of radius R_(S), called the 'Schwarzschild Radius'. It is the critical radius at which escape speed equals the speed of light c. Nothing even the light, can escape from within the sphere of Radius R_(S). So light from a black hole can not escape and hence the terminology 'black hole'. There has been astronomical evidence of a small and massive object at the centre of our galaxy the 'Milky way'. Suppose that there is a particle at a distance about 6 light years. that orbits this massive object with an orbital speed of about 2 xx 10^(5) m//s. Use the given data whereever necessary and answer the questions that hollow. G = 6.67 xx 10^(-11) N - m^(2)//kg^(2), Solar mass M = 2 xx 10^(30) kg, C = 3 xx 10^(8) m//s, 1 light year = 9.5 xx 10^(15) m. Theories suggest that it is not possible for a single star to have a mass of more than 50 masses. The massive object at the centre of milky way galaxy is most likely to be a : |
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Answer» WHITE dwarf |
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| 31. |
A planet of mass M moves around the sun along an ellipse so that its minimum distance from the sun is equal to r and the maximum distance to R. Making use of K epler's law, find the its period of revolution around the sun. |
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Answer» |
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| 32. |
A bimetallic strip consiting of a brass strip and a steel strip, each of length 1m and thickness 1//2 cm is clamped at one end at 0^(@)C.Calculate the depression of the free end when it is heated to 100^(@)C. |
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| 33. |
If two particles are moving on same circle or different circles in same plane with different angular velocities omega_(1) and omega_(2) and different time period T_(1), and T_(2) in same sense, then the time taken by 2 to complete one revolution w.r.t particle l is |
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Answer» `T= (T_(1)T_(2))/(T_(2)-T_(1))` |
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| 34. |
A uniform cube of mass m and side a is placed on a frictionless horizontal surface. A vertical force Fis applied to the edge as shown in figure. Match the following (most appropriate choice). |
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Answer» Solution :Torque of force about point A is `tau=Fa` (anticlockwise) Torque of weight mg of cube about point A `tau_(2)=mg((a)/(2))` (CLOCKWISE) The cube will not exhibit just rotational MOTION if `tau_(1)=tau_(2)` or `F=(mg)/(2)` Cube will rotate if `tau_(1) gt tau_(2)or F gt (mg)/(2)` For normal reaction force acting effectively at a DISTANCE `(a)/(3)` from point A, the cube will just rotate if `mgxx(a)/(3)=Fa` or `F=(mg)/(3)` When `F=(mg)/(4) LT (mg)/(3)`, there will be no rotational motion So, we conclude that `(a) to (ii), (b) to (III), (c) to (i), d to (iv)` |
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| 35. |
Take the effect of bulging of earth and its rotation is account . Consider the following statements. (A)There are points outside the earth where the value of g is equal to its value at the equator (B) There are points outside the earth where the value of g is equal to its value at the poles |
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Answer» Both A and B are CORRECT |
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| 36. |
(A) : A car can run on road because of force applied by road on the car.(R ) : Friction provides the necessary force for translatory motion for a car starting from rest. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 37. |
A force is appllied for a duration of 10 sec on a body of mass 5kg that is at rest . As a result the body acquires a velocity of 2ms^(-1). Find the magnitude of the force applied. |
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| 38. |
A car travels starting form rest with constant acceleration alpha and reaches a maximum velocity V. It travels with maximum velocity for some time and retards uniformly at the rate of beta and the to rest. If s is the total distance and t is the total time of journey then t = |
Answer» Solution : `V=0+ALPHAT,,t_(1)=(v)/(ALPHA)` `S_(1)=(1)/(2)alphat_(1)^(2)=(v^(2))/(2alpha)` `O-V=betat_(3),t_(3)=(v)/(beta)` `O^(2)-V^(2)=-2betaS_(3)` `S_(3)=(v_(2))/(2beta)=(1)/(2)betat_(3)^(2)` `t_(2)=(S_(2))/(V)=(S-(S_(1)+S_(3)))/(V)=(S-(v^(2))/(2)((1)/(alpha)+(1)/(beta)))/(V)` `t=t_(1)+t_(2)+t_(3)` `=(v)/(alpha)+(s)/(v)-(v)/(2alpha)-(v)/(2beta)+(v)/(beta) t=(s)/(v)+(v)/(2)((alpha+beta)/(alphabeta))` `s_(2)=s-(s_(1)+s_(2))=s-((v^(2))/(2alpha)+(v^(2))/(2alpha))` `=s-(v^(2))/(alpha)=s-(vt-s)` `s_(2)=2s-vt` [`S_(2)` = distance TRAVELLED with CONSTANT velocity] |
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| 39. |
It is easier to pull a lawn mower than to push it . Why ? |
| Answer» Solution :WHILL pusing,the vertical component of the applied force increase the effective weight of the lawn mover and hence the normal REACTION.So the sorce of friction is increased and it is difficult to move it.On the there hand when we pull a lawn mower, the vertical component of the applied force REDUCE is effective weight and normal reaction. conseqently, friction forces much REDUCED. | |
| 40. |
A gas consisting of diatomic molecules is at temperature of 117°C. if the moment of inertia of each molecule about an axis passing through the centre and normal to the molecular axis is 1.5 xx 10^(-40) "kgm"^(2). then the root mean square angular speed of each molecule is: (Boltzmann's constant k=1.38 xx 10^(-21) J//K ) |
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Answer» `4.5 xx 10^(9) "ras" s^(-1)` |
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| 41. |
A light open rigid paper frame as shown in (Fig.) floats on the surface of water. What will happen to the frame if some soap solution is dropped inside it ? What force will acts on the frame and in what direction will it act ? |
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Answer» `t_(h) = (T_(UP)h)/(l)` Water is kept in the capillary by surface tension. If `sigma_(h)` is the surface tension at the temperatrue `T_(h)`. We can write `h = (2sigma_(h))/(rho_(w)gr)`, Where `rho_(w)` is the density of water. Hence we obtain `sigma_(h) = (rhogrh)/(2) = ((rhogrl)/(2))((T_(h))/(T_(up)))`. Using the HINT in the CONDITIONS of the problem, we plot the graph of the funcation `sigma(T)`. The temperature `T_(h)`b on the level of the maximum ascent of water is determined by the point of intersection of the curves DESCRIBING the `(rhogl//2)T//T_(up)` and `simga(T)` dependences.  Figure shows that `T_(h) = 80^(@)C`. CONSEQUENTLY. `h = (lT_(n))/(t_(up)) = 6.4 cm`. Teh problem can also be solved ANALYTICALLY if we note that the `simga(T)` dependence is practically linear. |
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| 42. |
When a body is whirled in a vertical circle such that its total mechanical energy is constant with critical velocities then tension in the string changes with time. Arrange the following in increase order of tension when (i) String is horizontal(ii) String makes and angle 0^(@) with downward vertical(iii) String makes an angle of 180^(@) with downward vertical (iv) String makes an angle 120^(@) with downward vertical. |
| Answer» Answer :A | |
| 43. |
One mole of a gas is put under a weightless piston of a vertical cylinder at temperature T. The spaceover the piston opens into atmosphere. Initially piston was in equilibrium, How much work should be performed by some external force to increase isothermally the volume under the piston to twice the volume? (Neglect friction of piston) |
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| 44. |
A body of mass 300 kg is moved through 10m along a smooth inclined plane of angle 30^(@). The work done in moving is (jouls) (g= 9.8 m"/"s^2). |
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Answer» `4900` |
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| 45. |
What determines the natural frequency ofbody ? |
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Answer» Solution :NATURAL FREQUENCY of a body is determined by (i) Elastic PROPERTIES of the material of the body. (ii) DIMENSIONS of the body. |
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| 46. |
From a complete ring of mass .m. and radius .R. a 30^(@) sector is removed. The M.I. of the remaing ring about an axis through centre of ring and normal to its plane is |
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Answer» `(3)/(4)MR^(2)` |
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| 47. |
Is it unreasonable to expect a coefficient of friction to exceed unity ? |
| Answer» SOLUTION :In the CASE of ordinary plane surfaces the coefficient of friction is LESS than ONE. Suppose the surface is such that it is so irregular and there are very minute projection and cavities on it, then the coefficient of friction may exceed UNITY. | |
| 48. |
An object is dropped is an unknown planet from height 50 m, it reaches the ground is 2 s. The acceleration due to gravity in this unknwon planet is |
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Answer» `20MS^(-2)` `a=(2s)/(t^(2))=(2xx50)/((2)^(2))=25ms^(-2)`. |
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| 49. |
(A) : A needle placed carefully on the surface of water may float, where as aball of same material sinks (R ) : Archimedes principle is applicable for only spherical bodies |
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Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) |
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