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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. |
How does the accelerator increase the speed of a car? |
| Answer» Solution :The acceleration on being PRESSED increases the angular velocity of the wheel and thereby the rolling friction increases. At the point of contact with the ground the wheel ROTATES backwards so the rolling friction ACTS in the forward direction. The car accelerates DUE to the increase of this rolling friction. | |
| 2. |
The correct order in which the ratio fo SI unit to CGS unit increases is (a) Power (b) Surface tension (c ) Pressure (d) Force |
| Answer» Answer :A | |
| 3. |
A particle moves according to the law X =a cos (pi t)/2 . The distance covered by it in the time interval between t = 0 to 1 = 3 sec is |
| Answer» ANSWER :B | |
| 4. |
Show that the minimum speed at dle lowest point as sqrt(5gr)in a vertical circle executed by the object. |
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Answer» SOLUTION :To have this minimum speed `(v_2 = sqrt(gr) )`at point 2, the body must have minimum speed also at point 1. By making use of equation we can find the minimum speed at point 2 . ` v_1^2 - v_2^2 = 4gr` SUBSTITUTING equation` v_2 = sqrt(gr) " in " v_1^2 - v_2^2 = 4gr ` `v_1^2 - gr = 4gr` ` v_1^2 = 5gr` ` v_1 = sqrt(5gr)`....(2)  The body must have a speed at point 1, `v_1gt = sqrt(5gr)`to STAY in the CIRCULAR path. From equations `v_2 = sqrt(gr)`and `v_1= sqrt(5gr)`, it is clear that the minimum speed at the LOWEST point 1 should be 5 times more than the minimum speed at the highest point 2, so that the body loops without leaving the circle. |
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| 5. |
A particle of mass 'm' is projected with a velocity 'u' at an angle theta with the horizontal. Work done by gravity during its descent from its highest point to a point which is at half the maximum height is |
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Answer» `0` |
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| 6. |
A light string passing over a smooth light pulley connects two blocks of masses m_1 and m_2(vertically). If the acceleration of the system is g/8 then the ratio of the masses is |
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Answer» `8:1` ` m_2a = T -m_2 g"" [a= g/8]` `m_1 a = m_1 g - T` adding we get ` (m_1 + m_2) a = (m_1 - m_2) g ` PUTTING the VALUES, we get `m_1/m_2 = 9/7 ` |
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| 7. |
Two cylidrical vessels of equal cross sectional area contain water up to heights h_(1)(gth_(2)) and h_(2). The vessels are inter connected so that the levels in them become equal. Calculate the work done by the force of gravity during this process if rho is the density of water |
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Answer» `rhoA((h_(1)-H^(2))/2)^(2)G` |
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| 8. |
A body A of mass 0.5kg and specific heat 0.85J g^(-1) K^(-1) is at a temperature of 60^(@)C. Another body B of mass 0.3kg and specific heat 0.9 J g^(-1) K^(-10 is at a temperature of 90^(@)C. When they are connected through a conducting rod, heat will flow from |
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Answer» A to B `m_(B) = 0.3kg, s_(B) = 0.9 J g^(-1) K^(-1) and T_(B) = 90^(@)C` Heat always flows from a body of heigher temperature to a body of LOWER temperature. Since the body B is at a higher temperature, therefore heat will flow from body B to A |
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| 9. |
The distance of Neptune and Saturn from the sun are respectively 10^13 and 10^12 metres and their periodic times are respectively T_N and T_S If their orbits are assumed to be circular, then the value of T_N//T_S. |
| Answer» Answer :B | |
| 10. |
A body of mass 10 kg is raised from certain depth. By the time is raised by 10m, if its velocity is 2ms^(-1), work done during this time is, |
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Answer» 980 j |
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| 11. |
Calculate the colour temperautre of the sun assuming that the wavelength of maximum energy in the solar spectrum is 0.48 micron and the Wien's constant is 0.228xx10^(-2) mK. |
| Answer» Solution :Temperature of sun = `T =b/lamda_(m)=(0.288xx10^(-2))/0.48`= 6000 K | |
| 12. |
A particle starts with initial velocity 4m/s and moves such that acceleration is given by a = sqrt(V) It's velocity aftera time interval of 4 sec is |
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Answer» 20 m/s |
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| 13. |
A particle starts with initial velocity 4m/s and moves such that acceleration is given by a = sqrt(V) It's velocity after a displacement of (38)/(3) m is |
| Answer» Answer :A | |
| 14. |
If 'R' is the radius of the earth and 'w' is angular velocity with which it rotates about its axis, the variation in the value of 'g' at 45^(@) latitude of the earth stops its rotation will be |
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Answer» INCREASE by `(R W^(2))/(2)` |
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| 15. |
The velocity versus time of two particles moving along x-axis varies as shown in the following two plots. Then |
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Answer» maximum SEPARATION between the two particles is 2m |
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| 16. |
The breaking stress of steel is 7.9xx10^9 Nm^(-2) and density of steel is 7.9xx10^3 kgm^(-3)and g = 10ms^(-2) . The maximum length of steel wire that can hang vertically without breaking is |
| Answer» Answer :4 | |
| 17. |
As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m//s^(2)) |
Answer» Solution :Consider the equilibrium of each side of the ladder separately.  TENSION = `98 N, N_(B) = 245 N, N_(C) = 147 N`. |
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| 18. |
When a body moves with constant speed in a circular path, then ……………….. |
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Answer» WORKDONE will be zero |
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| 19. |
A simple pendulum is fixed on a trolly which is slinding down on a frictionless plance at an angle 60^(@) . The ratio of time period when the trolly is on a horizontal plane to the time period when the trolly is sliding is |
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Answer» `SQRT(3):2` |
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| 20. |
A particle starts moving from rest with uniform acceleration. It travels a distance x in the first 2 sec and a distance y in the next 2 sec. Then |
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Answer» `y=x` |
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| 21. |
Keeping the density of the earth constant, if the radius of the earth decreases by 2% , what would be thechange in ‘g.? |
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Answer» Solution : Relation between acceleration DUE to gravity g and radius of earth (R) in terms of density is `g = 4/3 pi Grho. R` When density is kept CONSTANT and radius is DECREASED by 2% then `(Delta R)/(R) XX 100 = 2` from distribution of errors in multiplications and divisions `(Delta g)/(g) xx 100 = (Delta R)/(R) xx 100 = 2%` When density is constant and radius of earth decreases by 2% then decrease in g value is also 2%. |
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| 22. |
The pressure of a gas of 5 litre volume at 20^(@)C is 10 atm. When the temperature of the gas is raised to 100^(@)C, its volume increased to 20 litres. What is the pressure of the gas at this temperature ? |
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Answer» |
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| 23. |
A satellite is revolving around earth its equational plane with a period T. If the radius of earth suddenly to half its radius without change in the mass. Then the new period of revolution will be |
| Answer» ANSWER :D | |
| 24. |
A body is thrown with velocity (4i + 3j) metre per seocnd. Its maximum height is (g = 10 ms^(-2)) |
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Answer» 2.5 m |
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| 25. |
In the obsence of wind the range and maximum height of a projectile were R and H. If wind imparts a horizontal acceleration a = g/4 to the projectile then find the maximum range and maximum height. |
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Answer» Solution :`H^(1) = H(:. u sin theta "REMAINS same"), T^(1) = T` `R^(1) = u_(x) T + (1)/(2)a T^(2) = R+(1)/(2)(g)/(4)T^(2) = R+(1)/(8) gT^(2) = R+H` 
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| 26. |
A body is thrown upward at intial velocity of 7 ms^(-1) , then at which height it's kinetic energy becomes half ? |
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Answer» Solution :Suppose at height h its KINETIC becomeshalveto its initial kinetic energy . `K = 1/2 (K_(0))/2 ` `1/2 mv^(2) = 1/2 (1/2mv_(0)^(2))` ` :. V^(2) = 1/2 v_(0)^(2) = 2gh""…(1)` but equation of a body thrown in upward direction . `v_(2)^(2) -v_(0)^(3) =2gh` `1/2v_(0)^(2) =v_(0)^(2) = - 2gh ` [ From equation (1)] ` :. -1/2 v_(0)^(2) = - 2gh` ` :. 2gh = 1/2 v_(0)^(2)` ` :. h = (v_(0)^(2))/(4G) = 49/(4xx9.8)` ` :. h = 1.25 m ` Alternative METHOD : From conservation of mechanical energy `E = K+V` ` :. ` Constant ` = K+V` For at point on ground and at point on height h `K_(1)+V_(1)=K_(2)+V_(2)` `:. K_(1) + 0 = (K_(1))/2 +V_(2)` ` :. V_(2) = K_(1) - (K_(1))/2 ` ` :. V_(2) = (k_(1))/2` ` :."mgh " = 1/2 (1/2 mv_(0)^(2))` ` :. h = (v_(0)^(2))/(4g)` `((7)^(2))/(4xx9.8)` ` :. h = 1.25 m ` |
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| 27. |
A bob is hung from an elastic spring to a support and set in motion in a vertical plane in a way similar to the oscillations of a simple pendulum. Its length will be maximum |
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Answer» At its LEFT EXTREME position |
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| 28. |
For a projectile the ratio of maximum height reached to the square of flight time is (g = 10 ms^(-2)) |
| Answer» ANSWER :A | |
| 29. |
Column-1 shows certain siruations and column-2 shows information about forces. |
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Answer» |
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| 30. |
What is the value of Reynolds number for unsteady flow ? |
| Answer» SOLUTION :Between 1000 to 2000. | |
| 31. |
A flexible but inextensible chain of length l is placed on a smooth table with an initial length 1 = nl hanging down the table. Calcultae the velocity with which the chain will leave the table if released from rest. Consider two cases : (i) the hanging part does not touch the ground whichis at a large distance, (ii) the lower end of the hanging part initially just touches the ground. Also calculate the time in which the chain leaves the table in the first case [Hint : Taking the table as the datum and m as the mass per unit length of the chain, apply the principle of conservation of energy]. |
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Answer» |
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| 32. |
A ball of radius r is made to oscillate in a bowl of radius R. Find the time period of oscillation. |
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Answer» Solution :The EFFECTIVE LENGTH of the simple PENDULUM L = (R - r) Time peviod T = `2 pi sqrt [(R-r)/g]` |
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| 33. |
A particle is moving in a circle of radius r under the action of a force F = alpha t^(2)which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy = 0 for r - 0): |
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Answer» `1/2 ALPHA t^(3)` |
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| 34. |
The blades of a windmill sweep out a circle of area A. What is the kinetic energy of the air? |
| Answer» SOLUTION :`K= PAV^(3)t"/"2` | |
| 35. |
Identity the correct statement (s) |
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Answer» The RELATION between potential difference and GRAVITATIONAL and field intensity `(vecE)` is`V_(A)- V_(B) = -int_(B)^(A)vecE.vec(dl)` |
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| 37. |
A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm. the block just loose contact with the plank when the plank is at momentary rest. Then |
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Answer» the period of oscillation is `((2pi)/(5))` |
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| 38. |
Length of a simple pendulum is1m. When its bob is at its lowest point its velocity is 7ms^(-1). If the bob leaves its circular path at a height .h. above the center of the circle, then the value of .h. is (g = 10 ms^(-2)) |
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Answer» `0.95 m` |
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| 39. |
If the velocity isvec(v) = 2 hat(i) + t^(2) hat(j) - 9 vec(k),then themagnitude of acceleration att = 0 . 5 sis: |
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Answer» `1ms^(-2)` at ` t= 0.5s impliesa= 2( 0.5 ) =1 ms^(-2)` |
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| 40. |
A body weighs 700gm-wt on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/7 and radius half of the earth |
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Answer» 200GM wt |
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| 41. |
The thermal conductivites of copper, mercury and glass are respectively K_(C),K_(M)andK_(G) such that K_(C)gtK_(M)gtK_(G). If the same quantity of heat flows per second per unit area of each and the corresponding temperature gradients are X_(C),X_(M)andX_(G) then |
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Answer» `X_(C)=X_(m)=X_(g)` |
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| 42. |
If I_(1) is the moment of inertia of a thin' rod about an axis perpendicular to its length and passing through the centre of mass and I_(2) is the moment of inertia of the ring formed by bending this rod in the form of a ring, then (I_1)/(I_2) is |
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Answer» `1 : 1` `therefore I_(1) = (1)/(12) ML^(2) ""because L = 2pi r or r = (L)/(2pi)` `therefore I_(2) = MR^(2) = M ((L)/(2pi))^(2)` THUS , `(I_(1))/(I_(2)) = (pi^2)/(3)` |
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| 43. |
A particle moves in the x-y plane under the action of a force vec(F) such that the componentsof its linear momentum vec(P) at any time t are p_(x) = cos t and p_(y) = 3 sin t. What is the magnitude of the vector vec(F)? |
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Answer» `2SQRT2` |
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| 44. |
The Brownian motion is due to: |
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Answer» The motion of molecules in FLUID medium |
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| 45. |
The distance travelled in a time t, by a particle moving along X-axis according to an equation x = A cos omega t is given by S = nA + S_0, when time t can be written as t = n(T/4) + t_0 such that t_0 < T/4 and distance travelled in that t_0 is S_0. The distance S is, when ‘n’ is even number |
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Answer» `S = A [(N + 1) - COS (wt + (n pi)/(2))]` |
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| 46. |
The distance travelled in a time t, by a particle moving along X-axis according to an equation x = A cos omega t is given by S = nA + S_0, when time t can be written as t = n(T/4) + t_0 such that t_0 < T/4 and distance travelled in that t_0 is S_0. The distance S is, when ‘n’ is odd number |
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Answer» `S = A [(N + 1) - cos (WT + (n PI)/(2))]` |
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| 47. |
Identify the lector quantity |
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Answer» PATH length |
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| 48. |
Rain is falling vertically with a speed of 20 ms^(-1). A person is running in the rain with a velocity of 5ms^(-1) and a wind is also blowing with a speed of 15 ms^(-1) (both from the west). The angle with the vertical at which the person hold his umbrella so that he may not get drenched is : |
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Answer» Solution :`VEC(V_("Rain")) = vec(V_(R)) = 20(-hat(K))` `vec(V_("Man")) = vec(V_(W)) = 15 hat(i)` Resultant velocity of rain and WIND = `vec(V_(RM)) = -20 hat(K) + 15 hat(i)` Now, Velocity of Rain relative to man = `vec(V_(RM)) - vec(V_(M))` `= (-20 hat(K) + 15 hat(i)) - (5 hat(i))` `= -20 hat(K) + 10 hat(i)` `Tan alpha = (1)/(2) rArr alpha = Tan^(-1)""(1)/(2)` 
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| 49. |
Define beats. |
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Answer» Solution :FORMATION of beats: When two or more waves superimpose cach other with different slightly different frequencies, then a sound of periodically varying amplitude at a point is OBSERVED. This phenomenon is KNOWN as beats. The number of amnplitude maxima per second is called frequency. If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second `n=|f_(1)-f_(1)| "per second"` |
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| 50. |
In a cylinder-piston arrangement air is under a pressure P_(1) and constant temperature t. A soap bubble lies insie the cylinder, surface tension is T. The soap bubble is to be reduced to half its size, the required pressure P_(2) to which air should be compressed by slowly pulling the piston into the cylinder is |
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Answer» `8P_(1)+(24T)/(R)` |
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