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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. |
In a cylindrical vessel containing liquid of density rho, there are two holes in the side walls at heights of h_1 and h_2 respectively such that the range of efflux at the bottom of the vessel is same. The height of a hole, for which the range of efflux would be maximum, will be – |
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Answer» `h_(2)-H_(1)` |
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| 2. |
Give three examples for a quantity which has unit, but still dimensionless. |
| Answer» SOLUTION :ANGLE (UNIT RADIAN) | |
| 3. |
Find the angle between force F= (3hati+4hatj-5hatk) unit and displacement d= (5hati+4hatj+3hatk) unit. Also find the projection of F and d. |
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Answer» Solution :`F*d = F_(x)d_(x)+F_(y)d_(y)+F_(Z)d_(z)` `=3(5)+4(4)+(-5)(3)= 16` unit. Hence `F*d= Fd cos THETA= 16` unit Now `F*F= F^(2)= F_(x)^(2)+F_(y)^(2)+F_(z)^(2)` `=9+1625= 50` unit and `d*d= d^(2)= d_(x)^(2)+d_(y)^(2)+d_(z)^(2)` `=25+16+9= 50` unit `cos theta= (16)/(SQRT(50)sqrt(50))= (16)/(50)= 0.32` `theta = cos^(-1) 0.32`. |
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| 4. |
The ratio of intensities of radiation at distances of R and 3R from a spherical source will be |
| Answer» ANSWER :B | |
| 5. |
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 :K.E=100 J,`n_(1)`=60 rpm =1 RPS,`n_(2)`=180 rpm =3rps Increase in kinetic energy `E=(1)/(2)I(omega_(2)^(2)-omega_(1)^(2))=(1)/(2)I(4pi^(2)n_(2)^(2)-4pi^(2)n_(1)^(2))` `100=(1)/(2)I(4pi^(2).9-4 pi^(2)1^(2))=2pi^(2)I(9-1)=16pi^(2).I` `therefore` momeny of inertia of the wheel I =`(100)/(16 mpi^(2))=0.633 kg m^(2)` |
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| 6. |
Air is initially at 260^@ C and 700 N/m^2 and occupied 0.028 m^3. The air is expanded at constant pressure to 0.084 m^3. A polytropic process with x = 1.5 is then carried out followed by a constant temperature process which completes a cycle. (a) Sketch P-V graph (b) Find heat received and heat rejected in the cycle (c) Efficiency of cycle |
| Answer» SOLUTION :`124J,-17.7 J,-58.4 J(C )38.6 %` | |
| 7. |
If L = (20 pm 0.01) m and B = (10 pm 0.02)m then L/B is |
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Answer» `(2 +- 0.03)m` |
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| 8. |
Parsecis unit of which physical quantity ? |
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Answer» TIME |
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| 9. |
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds How manyrevolutions does the engine make during this time ? |
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Answer» 376 |
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| 10. |
A constant torque of 1000 Nm turns a wheel of M.I 200 kg m^(2) about an axis through centre. The angular velocity after 3s is |
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Answer» 15 RAD `s^(-1)` |
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| 11. |
In which condition ,work done on the body does nit increase its kinetic energy ? |
| Answer» Solution :When BODY is moving as a FRACTIONAL surface , GIVEN work is USED to oppose the FRICTIONAL force . | |
| 12. |
Examine Tables 6.1-6.3 and express The kinetic energy of an air molecule (10^(-21)J) in eV. |
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Answer» SOLUTION :The KINETIC energy of an air molecule is `(10^(-21)J)/(1.6xx10^(-19)J"/"eV) cong 0.0062 eV` This is the same as 6.2 MEV. |
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| 13. |
The angular amplitude of simple pendulum is theta_0. The maximum tension in its string will be (assume theta_0 is small) |
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Answer» `MG (1 + theta_0)` |
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| 14. |
Let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), themass of the object is 3 kg, spring constant is 1200 N/m and the amplitude is 2 cm. (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase? |
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Answer» Solution :(a) x = 2 sin 20T (b) x = 2 cos 20t (C) x = - 2 cos 20t where x is in cm. These functions DIFFER neither in amplitude nor frequency. They differ in initial PHASE. |
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| 15. |
A metal ring is'melted and a solid sphere is made out of it. What happens to the M.I. about a vertical axis through the centre ? |
| Answer» Solution :M.I. DECREASES DUE to decrease in distance of the redistributed,MASS. | |
| 16. |
A brass boiler has a base area of 0.15m^(2) and thickness 1 cm. Water boils in it at the rate of 6.0 kg/min, when placed on flame. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J/sm-K and latent heat of vapourisation of water = 2256 J/g. |
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Answer» Solution :Let `T_(1)` be the temperature of the part of the flame in contact with BOILER. The amount of heat flown into water in 1 MINUTE `Q=(KA(theta_(1)-theta_(2))t)/L=(109xx0.15xx(T_(1)-100)xx60)/0.01` Mass of water boiled in 1 minute = 6KG = 6000g. Heat required to boil the water, Q = mL = `6000xx2256` `therefore(109xx0.15xx(T_(1)-100)xx60)/0.01=6000xx2256` (or) `T_(1)-100=138""thereforeT_(1)=238^(@)C` |
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| 17. |
The free end of a simple pendulum is attached to the ceiling of a box. The box is taken to a height and the pendulum is oscillated. When the bob is at its lowest point the box is released to fail freely. As seen from the box during this period, the bob will |
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Answer» CONTINUE its OSCILLATION as before |
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| 18. |
A lead bullet of mass 12 g travelling at a speedof 100 ms^(-1)comes to rest in a wooden block. If no heatis taken away by the wood, find the rise in temperature of the wood. (Specific heat of lead = 0.03 calorie/g ""^(@)C.) |
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Answer» Solution :kineticenergy of the bullet= heatgained by the bullet `(1)/(2) mv^(2) = MS Delta t` `Delta t =(v^(2))/(2S) = ((100)^(2))/( 2xx0.03 XX 4.2 xx 1000) = 39.68^(0)C` |
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| 19. |
70 cal of heat is required to rise the temperature of 2 moles of an ideal gas at constant pressure from 30^@C to 35^@C . What is the amount of heat required to rise the temperature of same gas through the same range al constant volume? (R = 2cal/mole/K) |
| Answer» ANSWER :B | |
| 20. |
The optical path of a monochromatic light is the same if it goes through 4.00m of glass are 4.50 m of a liquid . If the refractive index of glass is 1.5, what is the refractive index of the liquid? |
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Answer» Solution :When LIGHT travells a DISTANCE x in a medium of refractive INDEX `mu` the optical PATH is `mu x` Given `mu_x x_1= mu_2 x_2 implies 1.5 times 4.00=mu_2 times 4.50` `mu_2=(1.5 times 4.00)/4.50=1.333` |
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| 21. |
The velocity of a body of mass 2 kg as a function of t is given by v(t) = 2t hati + t^(2)hatj . Find the momentum and the force acting on it, at time t = 2s. |
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Answer» Solution :MASSOF the bodym=2 kg VELOCITYOF thebodyv (t ) = (2t )`hat(j)+ (t^(2) hat(j))` Velocityof thebodyat t= 2 s `=4hat(j ) + 4hat(j )` Momentumof body`VEC( p )=mv= 2 (4hat(j ) + 4hat(j ))` Acceleration of body `a= (d v)/( dt )` `=(d )/(dt ) (2t hat(j ) + t^(2) hat(j ))` Att =2s `(2HAT(i ) + 4hat(j ) )` FORCE acting on the body( F ) =ma `=(4hat(i )+ 8hat(j ))N` |
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| 22. |
A lead bullet of initial temperature 27^(@)C and speed 'V' kmph pnetrates into a solid object and melts. If 50% of the kinetic energy is used to heat it, the value of v in kmph is (for leads melting point=600K, latent heat of fusion=2.5xx10^(4)JKg^(-1), "specific heat" =125 J kg^(-1) k^(-1)) |
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Answer» 1800 |
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| 23. |
A block of mass 2.5 kg is pushed 2.2. m along a frictionless horizontal table by a constant 16 N force directed 45^(@) above the horizontal. Then Workdone by the force of gravity |
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Answer» 24.9J |
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| 24. |
Displacement (in metre) of a particle varies with time (in second) is given as y(t)= 4t ^(2) - 16t +5.Time taken by the particle to come to rest is……. |
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Answer» Solution :`y (t) = 4T ^(2) - 16 t + 5` `therefore v (t) = (d)/(DT) [ 4t ^(2) - 16 t +5]` `therefore v (t) = 8t - 16` If `v (t) =0 ` then time is .t. `therefore 0= 8t. -16` `therefore t.=2` second |
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| 25. |
A block of mass 2.5 kg is pushed 2.2. m along a frictionless horizontal table by a constant 16 N force directed 45^(@) above the horizontal. Then Workdone by the applied force is |
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Answer» 24.9J |
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| 26. |
The speed of a projectile at its maximum height is (sqrt(3))/(2) times its initial speed. If the range of the projectile is p times the maximum height attained by it, then p = |
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Answer» `4//3` |
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| 27. |
In the arrangement shown in figure the ends of an in extensible string move downwards with uniform speed u. Pulleys A and B are fixed. Find the speed with which the mass Mmoves upwards. |
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Answer» Solution :Suppose the distance of point O from the ceiling is y and the distance of point O from each PULLEY is x and the distance between the TWO pulleys is l. `x^(2) = y^(2) + (l^(2))/(4)` Differentiating the above equation w.r.t to "t" `2x((dx)/(d t)) = 2y((dy)/(dt)) + (1)/(4) 2l((DL)/(d t))` But, `u = -(dx)/(dt), v = (dy)/(dt)` and `(dl)/(d t)` `therefore - x u= y u` and `v = -u(x)/(y) = u sec theta` `therefore` Velocity of mass `= v = u sec theta` (UPWARDS) 
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| 28. |
An addiditional bridge is kept below a sonometer wire so that it is divided into two segments of lengths in the ratio 2 : 3 and n_(1), n_(2) are their respective fundamental frequencies. If the additional bridge is removed then the fundamental frequency of that sonometer wire is n, the ratio of n,n_(1),n_(2) is |
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Answer» `2 : 3 : 5` |
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| 29. |
Refer to figure. Let and be the change in internal energy in processes A and B respectively, DeltaQ be the net heat given to the system in process A+B and DeltaW be the net work done by the system in the process A+B a) DeltaU_(1)+DeltaU_(2)=0 b) DeltaU_(1)-DeltaU_(2)=0 c) DeltaQ-DeltaW=0 d) DeltaQ+DeltaW_=0 |
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Answer» Only a, B are CORRECT |
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| 30. |
What is the need of banking a circular raod |
| Answer» SOLUTION :Horizontalcomponentof the normalreactionof theroadprovides thenecessarycentripetalforcefor thevehicleto moveit ALONG thecurvedpath. | |
| 31. |
An external device, e.g., an electric motor, supplies constant power to a rotating system, e.g., a flywheel , through a torque tau. The angular velocity of the system of the system is omega. Both tau and omega are variable |
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Answer» `omega prop TAU` |
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| 32. |
Volume of some air saturated with water vapour is 80 cm^3 at the pressure of 74 cmHg .Keeping the temperature fixed, if pressure is taken to 146 cmHg, then the volume is halved. What is the pressure of the water vapour in this state? |
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Answer» |
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| 33. |
A body constrained to move in the z direction, is subjected to a force given by vecF = 2 hati +15hatj + 6hatk N. What is the work done by the force in moving the body a distance 10 m along the z-axis? |
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Answer» 20 J |
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| 34. |
A body is moving with a constant speed v in a circle of radius r. its angualr acceleration is : |
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Answer» VR |
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| 35. |
It is hotter over top of the fire than at same distance on the sides due to |
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Answer» heat is radiated UPWARDS |
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| 36. |
For a particle executing SHM. The kinetic energy 'K' is given by k = k_(0) Cos^(2)wt. The maximum value of KE is |
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Answer» `K_(0)` |
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| 37. |
A uniform rod of mass m and length l is suspended by means of two light inextensible strings at the rnds of rod. Tension in one stringimmediately after the other string is cut is |
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Answer» Solution :`mg-T=ma ""`…..(1) `alpha = (tau)/(I)=(mg(l)/(2))/((ml^(2))/(3))=(3G)/(21) ""`…..(2) `a=(l)/(2)alpha ""`….(3) Solving equations (1), (2) and (3) we GET, `T=(mg)/(4)` 
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| 38. |
Two masses m_(1) and m_(2) are suspended from a spring of spring constant 'K'. When the masses are in equilibrium, m_(1) is gently removed. The angular frequency and amplitude of oscillation of m_(2) will be respectively |
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Answer» `sqrt((k)/(m_(2))), (m_(1)G)/(k)` |
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| 39. |
The linear momentum of a particle as a function of time t is given by p = a + bt, where a and b are positive constants. The force acting on the particle is |
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Answer» a |
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| 40. |
A gas bubble of diameter 002 m rises steadily at the rate of 2.5 10^(-2) ms^(-1)through a solution of density 2.5 xx 10^(3)kg m ^(-3) . Calculate the coefficient of viscosity of the solution. Neglect the density of the gas. |
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Answer» |
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| 41. |
The Young's modulus for a perfect rigid body is |
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Answer» 0 |
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| 42. |
If earth shrinks to 1//64 of its volume with mass remaining same duration of day will be |
| Answer» ANSWER :A | |
| 43. |
The blades of a windmill sweep out a circle of area A . (a) If the wind flows at a velocity v perpendicular to the circle , what is the mass of the air passing through it in time t ? (b) What is the kinetic energy of the air ? (c ) Assume that the windmill converts 25 % of the wind's energy into electricalenergy , and that A = 30 m^(2) , v = 36 km/h and the density of air is 1.2 m^(2) v = 36 km/h and the density of air is 1.2 kg m^(-3) .What is the electrical power produced ? |
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Answer» Solution :(a) Suppose area of the circle swept by the windmill is A, velocity of the WIND is v and density of air = `rho ` Distance covered of air passing through circular area A , V = Ad ` :. V = Avt [ :. d = vt] ` ` :. ` MASS of flowing air = volume `xx` density m =` Avtrho` (b) Kinetic energy of air pasing through area of A in time t , `K = 1/2 mv^(2)` `=1/2 xxAvt rho xx v^(2)` ` K = 1/2 A rho vt^(3)` (c ) Input kinetic energy is energy of wind and output energy GENERATED , ` :. "EFFICIENCY " eta = (" Output energy")/("Input energy")` 25 % = `("ELECTRICAL energy ")/(1/2Av^(3))` ` :. ` Electric energy produced ` =1/4 xx1/2 xx 0 xx 1.2 xx(10)^(3) xxt` = 4500 t where `v = (36xx1000)/(3600) = 10 ms^(-1)` ` :. "Electric power " = (" Electric energy")/("Time")` = 4500 W `=4.5 W` |
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| 44. |
If the coefficient of friction between the tyres of a car and the road is mu_(k), show that the maximum stopping distance of the car when moving with a velocity v is v^(2)//2mu_(k)g |
| Answer» SOLUTION :`a=-F_(K)//m=-mu_(k)mg//m=-mu_(k)G`, `s=v^(2)//(-2)a=v^(2)//2mu_(k)g` | |
| 45. |
A circular metal disc of mass 200 g and radius 20 cm makes 10 revolutions per second about an axis passing through the centre and perpendicular to its plane. Find (a) Moment of inertia about the axis, (b) Torque required to change its angular velocity from 10 rps to 20 rps in5s. |
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Answer» SOLUTION :(a) `I=(Mr^(2))/(2)=(0.2xx(0.2)^(2))/(2)=4xx10^(-3)kgm^(2)`. (b) TORQUE `tau = I alpha = (I(omega_(2)-omega_(1)))/(t)` `therefore tau = (4xx10^(-3)(40 pi -20 pi))/(5), tau = 0.05 Nm`. |
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| 46. |
A particle moving in the positive x-direction has initial v_(0). The particle undergoes retardation kv^(2), where it its instantaneous velocity. The velocity of the particle as a function of time is given by |
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Answer» `v=v_(0)//(1+kv_(0)t)` |
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| 47. |
In a nuclear reactor a neutron of high speed (typically 10^(7) ms^(-1)) must be slowed to of interactingwith isotope._(92)^(235)U and causing it to fission . Show that a neutroncan lose mostof its kineticenergy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutronmass . the material makingup the light nuclei , usually heavy water (D_(2)O) or graphite is called a moderator . |
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Answer» Solution :The initial kinetic energy of the neutron is `K_(1i)= (1)/(2)m_(1)v_(1i)^(2)` While its final kinetic energy from Eq. (6.27) `K_(1f)=(1)/(2)m_(1)v_(1f)^(2)= (1)/(2)m_(1)((m_1 -m_2)/(m_1 +m_2))^(2)v_(1i)^(2)` The FRACTIONAL kinetic energy lost is `f_(1)=(K_(1f))/(K_(1i))=((m_1-m_2)/(m_1+m_2))^(2)` while the fractional kinetic energy gained by the moderating nuclei `K_(2f)"/"K_(1i)` is `f_(2)= 1- f_(1)` (elastic COLLISION) `=(4m_(1)m_(2))/((m_(1)+m_(2))^2)` One can also verify this result by substituting from Eq. (6. 28). For deuterium `m_(2)= 2m_(1)` and we obtain `f_(1)= 1"/"9" while "f_(2)= 8"/"9`. Almost 90% of the neutron.s energy is transferred to deuterium. For carbon `f_(1)= 71.6%" and "f_(2)= 28.4%`. In practice, howerver, this NUMBER is smaller since head-on-collisions are rare. |
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| 48. |
While determining the time period of oscillation of a simple pendulum, the readings from various measurements are 1.73s, 1.62s, 1.52s, 1.45s " and "1.83s. Calculate the values of mean value of time period, absolute error, mean absolute error and relative error. |
| Answer» SOLUTION :`1.63s, 0.12s, PM 0.074` | |
| 49. |
(A) : The force of attraction on a point mass situated inside a hollow spherical shell of uniform density is zero. (R) : When a point mass is placed inside a solid sphere, the vector sun of all the gravitational forces on it is zero |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 50. |
A cyclic process ABCD is shown in the p-V diagram. Which of the following curves represent the same process ? |
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Answer»
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