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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. |
the temperature of a gas is raised from 27^0 C to 927^0 C.The r.m.s molecular speed is |
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Answer» remains unchanged |
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| 2. |
A particle is moving east wards with a velocity of 5 ms^(-1)In 10 S , the velocity changes to 5 ms^(-1)north wards. The average acceleration in this time is |
Answer» Solution :`vec(a)_(a v) = (vec(DELTA v))/(Delta t) = (vec(v)_(f) - vec(v)_(i))/(Delta t)`  `vec(a)_(a v) = (5hat(j) - 5hat(j))/(10) = (5sqrt(2))/(10) = (1)/(sqrt(2))m//s^(2)` north-west direction. |
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| 3. |
Water at 20^(@)Cis flowing in a horizontal pipe that is 20.0m long.The flow is laminar and the water completely fills the pipe. A pump maintains a gauge pressure of 1400Pa, at a large tank at one end of the pipe. The other end of the pipe is open to the air, The viscosity of water at 20^(@)C is 1.005 poise (a) If the pipe has diameter 8.0 cm, what is the volume flow rate? (b) What gauge pressure must the pump provide to achieve the same volume flow rate for a pipe with a diameter of 4.0cm? (c) For pipe in part (a) and the same gauge pressure maintained by the pump, what does the volume flow rate become if the water is at a temperature of 60^(@)C (the viscosity of water at 60^(@)C is 0.469 poise? |
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Answer» `=((pi)/(8))((0.04)^(4))/(1.005xx10^(-1))xx(1400)/(0.2)` `=7xx10^(-2)m^(3)//s`. (b) `Q_(1)=Q_(2)` `:. [(p_(1)-p_(2))R^(4)]_(i)=[(p_(1)-p_(2))R^(4)]_(F)` Since, diameter or radius has decreased to half. Therefore gauge pressure should become 16 times or, `(p_(1)-p_(2))_(f)=16xx1400` `=2.24xx10^(4)Pa` (C )`Q prop (1)/(eta)` `:. (Q_1)/(Q_2)=(eta_2)/(eta_1)` or, `Q_(2)=((eta_(1))/(eta_(2)))Q_(1)` `=((1.005)/(0.469))(7xx10^(-2))` `=0.15m^(3)//s`. |
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| 4. |
The velocity of a particle change with time according to the relation v=xt+yt^(2)+z. Give the dimensions of x,y and z if v is in ms^(-1) and t is in s. |
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Answer» |
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| 5. |
Mass of a bus is 2400 kg . ……… J . Work is to be done in producing velocity 36 (km)/hin .it |
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Answer» `1.2 xx10^(6)` Now , Work `W = K.E .K =1/2 "mv"^(2) "" (:. "Taking " v_(0) = 0)` ` = 1/2 xx2400 xx10^(2)` `= 1.2 xx10^(5) J ` |
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| 6. |
Which of the following observations point to the equivalence of intertial and gravitational mass (a) Two spheres of difference masses dropped from the top of a long avacuated reach the bottom of the tube at the same time. (b) The time-period of a simple pendulum is independent os its mass. (c) The gravitational force on a particle inside a hollow under isolated is zero. (d)For a mss in a closed cabin that is falling frelly under gravity, gravity 'disappears'. (e) An astronautinside a sepaceship orbiting around the Earth fells weightless. (f) Planets orbiting around the sun obey kepler's third law (approximately). (g) The gravitational force on a body due to the Earth is equal and opposite to the gravitational force on the Earth due to the body. |
| Answer» SOLUTION :All axcept (C ) and (G). | |
| 7. |
Two satellites A and B are revolving around the earth in circular orbits of radius r_(1) and r_(2) respectively with r_(1)ltr_(2). Plane of motion of the two are same. At position 1, A is given an impulse in the direction of velocity by firing a rocket so that it follows an elliptical path to meet B at position 2 as shown. focal lengths of the elliptical path are r_(1) and r_(2) respectively. at position 2, A is given another impulse so that velocities of A and B at 2 become equal and the two move together. for any elliptical path of the satellite time period of revolution is given Kepler's planetry law as T^(2)alphar^(3) where a is semi-major axis of the ellipse which is (r_(1)+r_(2))/2 in this case. also angular momentum of any satellite revolving around the earth will remain a constant about earth's centre as force of gravity on the satellite which keeps it in elliptical path is given its position vector relative to the earth centre. If r_(2)=3r_(1) and time period of revolution for B be T then time taken by A in moving from position 1 to position 2 is |
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Answer» `T(sqrt(3))/(sqrt(2))` From `A` to `B`, field due to shell is zero, but fiedl due to SPHERE is non-zero. Hence, `t_(AB)ltR/(U_(A))lt(R^(2))/(sqrt(GMh))` |
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| 8. |
Two scooters start at an interval of 1 min between them, each moving with a uniform acceleration of 0.4 m//s^(2). How much later the distance between them would be 4.2 km? |
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Answer» 195s |
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| 9. |
Two identical particles move towards each other with velocity 2 v and v respectively. The velocity of centre of mass is |
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Answer» Solution :GIVEN `m_(1)=m_(2)=m` `V=2v` and `v_(2)=v` `:.v_(CM)=(m_(1)v_(1)+m_(2)v_(2))/(m+m)` `=v/2` |
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| 10. |
Two satellites A and B are revolving around the earth in circular orbits of radius r_(1) and r_(2) respectively with r_(1)ltr_(2). Plane of motion of the two are same. At position 1, A is given an impulse in the direction of velocity by firing a rocket so that it follows an elliptical path to meet B at position 2 as shown. focal lengths of the elliptical path are r_(1) and r_(2) respectively. at position 2, A is given another impulse so that velocities of A and B at 2 become equal and the two move together. for any elliptical path of the satellite time period of revolution is given Kepler's planetry law as T^(2)alphar^(3) where a is semi-major axis of the ellipse which is (r_(1)+r_(2))/2 in this case. also angular momentum of any satellite revolving around the earth will remain a constant about earth's centre as force of gravity on the satellite which keeps it in elliptical path is given its position vector relative to the earth centre. When A is given its first impulse at that moment |
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Answer» `A,B` and the earth centre are in same straight line Substuitute `V_(1)` and `r_(1)` and simplify, `E=-(GMm)/(2a)` |
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| 11. |
The densities of a liquid at 0^(@)C and 100^(@)C are respectively 1.0127 and 1. A specific gravity bottle is filled with 300 g of the liquid at 0^(@)C upto the brim and it is heated to 100^(@)C. Then, the mass of the liquid expelled in gram is Coefficient of linear expansion of glass = 9 xx 10^(-6)//^(@)C) |
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Answer» Solution :`d _(t) = (d _(0))/( 1 + gamma t) IMPLIES 1 + gammat = (d_(0))/( d _(t)) = 1. 018 implies gamma t = 0.018 implies gamma = (0.018)/(100) = 0.00018 per ^(@)C` `gamma _(APP) = 0.000180 -0.000027 = 0.000153 implies -000153 = (X)/((340 -x) xx 1000)` `x =0.0153 (340 -x) implies 1.0153 x = 340 xx 0.0153 implies x = (340 xx 0.0153)/( 1.0153) = 5.126 gm` |
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| 12. |
Meniscus of mercury in a capillary is ………………. . |
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Answer» concave |
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| 13. |
Find the moment of inertia of a sphere about a tangent to the sphere. Given the moment of inertia of the sphere about any of its diameters to be (2MR^(2))/(5) , where M is the mass of the sphere and R is the radius of the sphere. |
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Answer» Solution :The centre of mass (cm) of the sphere is on its diameter AB Fig .`I_(cm)=(2)/(5) "MR"^(2)` ACCORDING to the parallel-axes theorem the MOMENT of inertia of the sphere about the tangent CD, `I=I_(cm)+MR^(2)=(2)/(5)MR^(2)+MR^(2)=(7)/(5)MR^(2)` 
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| 14. |
Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 xx 10 ^(-3)kg. |
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Answer» Solution :We know that at STP, volume of ONE mole of gas is `V= 22.4 ` litre `=22.4 XX 10 ^(-3) m ^(3)` and pressure is `P =1 ` atm `= 1.01 xx 10 ^(5) (N)/(m^(2))` Density of air at STP, ` rho (M)/(V) = (29 xx 10 ^(-3))/(22.4 xx 10 ^(-3)) = 1.29 (kg)/(m^(3))` Approximate value of speed of sound in air at STP can be found out using following Newton.s formula, `v = sqrt ((P)/(rho)) = sqrt ((1.01 xx 10- ^(5))/(1.29)) =279.8ms ^(-1) =280 ms ^(-1)` Practicaly, speed of sound in air at STP is found to be `331 ms ^(-1).` Hence, percentage error in above approximation would be, `(331-280)/(331) xx100% =15.41%` Thus, here Newton.s formula gives speed of sound in air at STP, `15.41%` less than the experimental value. |
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| 15. |
A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed 0. A person of mass Mis standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is |
| Answer» Answer :B | |
| 16. |
Same heat is given to Helium and Oxygen gas of equal masses under constant volume. Then the temperature of the gas which increases more is |
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Answer» OXYGEN |
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| 17. |
What should be the lengths of steel and copper rod so that the length of steel rod is 10 cm longer than the copper rod at all the temperatures. Coefficients of linear expansion for copper and steel are alpha_(cu) = 1.7 xx 10^(-5)""^(0)C^(-1) and alpha_("steel") = 1.1 xx 10^(-5) " "^(0)C^(-1) |
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Answer» SOLUTION :`1_(ST) - 1_(cu) = 10CM - RARR (1)` `(Delta 1)_(St) = (Delta l)_(Cu)` `1_(St) prop_(St) Delta t 1_(Cu) prop_(Cu) Delta t rarr `(2) Solving (1) & (2) `1_(Cu) = 18.3`cm, `1_(st) = 28.3 ` cm |
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| 18. |
Two identical eggs, one raw and other boiled, area rotated with the same angular speed. Which one will come to rest earlier? |
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Answer» can.t say anything Now moment of inertia of raw egg `I_(1)gt` moment of inertia of boiled egg `(I_(2))`. both having same torque durin rotational motion so `I_(1)alpha_(1)=I_(2)alpha_(2)`, where `alpha_(1)andalpha_(2)` are anti angular accelerations of raw and boiled egg respectively. Here`I_(1)gtI_(2)` hence `alpha_(1)ltalpha_(2)` so raw egg has less anti angular ACCELERATION and hence it takes more time for rest, while `alpha_(2)` is more in boiled egg, so angular velocity slowly reducing to zero and it come to rest. |
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| 19. |
Explain the need for the measurements in Physics and briefly explain the measuring process of any physical quantity. |
| Answer» SOLUTION :1 (B) 1 and 1 (b) 2 | |
| 20. |
When a force is applied on a wire of uniform cross-sectional area 3×10−6 m2and length 4m, the increase in length is 1 mm. Energy stored in it will be (Y=2×1011 N/m2 ) |
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Answer» 6250J |
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| 21. |
A stone is projected from the ground with a velocity of 20 m/s at angle 30^(@) with the horizontal. After one second it clears a wall then find height of the wall. (g=10ms^(-2)) |
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Answer» |
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| 22. |
A constant force F acts on a moving body along a smooth quarter circular arc of radius R fixed on horizontal surface. Calculate the work done by force F from A to B |
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Answer» If force F always POINTS towards B then WORK `Fsqrt(2)R` |
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| 23. |
A uniform elastic plank moves over a smooth horizontal plane due to a constant force F, distributed uniformly over the end face. The surface of the end face is equal to A and Young.s modulus of the material is Y. Find the compressive strain of the plank in the direction of the acting force. |
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Answer» |
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| 24. |
A uniform metal rod of length L and mass M is rotating about an axis passing through one of the ends and perpendicular to the rod with angular speed omega. If th temperature increases by t^(@)C, then the change in its angular velocity is proportional to which of the following? (Coefficient of linear expansion of rod = a) |
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Answer» `SQRT(OMEGA)` |
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| 25. |
A closed gas cylinder is divided into two parts by a piston held tight. The pressure and volume ofgas in two parts (left and right) respectively are (P, 5V) and (10P, V). If now the piston is left free and the system undergoes isothermal process, then the volume of the gas in two parts (left and right) respectively are : |
| Answer» Answer :A | |
| 26. |
A solid sphere rolls without slipping on an inclined plane, then which of the following are correct ? |
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Answer» Torque of friction about point of contact is zero |
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| 27. |
A body is projected vertically up with a velocity of 60 m/s. The percentage of its initial kinetic energy converted into potential energy after 3 s is (g=10ms^(-2)) |
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Answer» 0.25 |
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| 28. |
If the length of the simple pendulum is increased by 44% from its original length, calculate the percentage increase in time period of the pendulum. |
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Answer» Solution :Since `Tpropsqrt(L)` Therefore, T = constant `sqrt(l)` `(T_(f))/(T_(i))=sqrt((l+(44)/(100)l)/(l))=sqrt(1.44)=1.2` Therefore, `T_(f)=1.2T_(i)=T_(i)+20%T_(i)` |
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| 29. |
A solid sphere of mass 2 kg is resting insidea cube as shown in the figure . The cube is moving with a velocity . Here t is in second . All surface are smooth . The sphere is at rest with respect to thecube . What is the totalforceexerted by the sphere on the cube . |
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Answer» Solution :`v=5ti + 2tj , bar(a) = (dv)/(dt) = 5i + 2J` ` F = ma_x HATI + m(G +a_y) hatj , F=2(5i) + 2[10+2]hatj` , `F=10i + 25 j , F= sqrt(100 + 567) = sqrt(676) = 26N` |
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| 30. |
A block of mass 1kg is dropped on a spring from a height of 20 cm. If the spring compresses by 5 cm, the force constant of the spring is (g = 10 ms^(-1)) |
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Answer» `2000 Nm^(-1)` |
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| 31. |
The weight of a balloon is W_1 when empty and W_2 when filled with air. Both are weighed in air by the same sensitive spring balance and under identical conditions. |
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Answer» `W_1 = W_2`, as the weight of air in the balloon is offset by the force of BUOYANCY on it. |
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| 32. |
If a circular piece of tin has a measured radius of 2.6 cm . What is circumference ? |
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Answer» Solution :`r=2.6cm` Circumference of circular dise = `2pir=2xx3.1428xx2.6` Here `2.6`has onley 2 significant DIGITS. Hence in the above multiplication `PI` VALUE should be written with `2+1=3` significant FIGURES. `pi=3.1428=3.14` Circumference `=2xx3.14xx2.6=16.328` This is to be rounded off to 2 significant digits. the result is 16. |
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| 33. |
a uniform circular disc of radiu8s r placed on a roughn horizontal plane has initial velocity v_(0) and an angul,ar velocity omega_(0) has shown The disc comes to rest after moving some distance in the direction of motion. Then |
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Answer» the friction force acting in the TOWARDS direction |
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| 34. |
A man crosses a river in a boat. If he crosses the river in minimum time he takes10 minutes with a drift 120m. If he crosses the river taking shortest path, he take 12.5 minute :( Assume v_(b//r) gt v_(r)) |
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Answer» WIDTH of the river is 200 m |
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| 35. |
The energy required to accelerate a body from 30 ms^(-1) " to" 60 ms^(-1) is three times the energy required to accelerate it from 10ms^(-1) to 'v'. The value of 'v' is |
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Answer» `30 MS^(-1)` |
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| 36. |
The root mean square speed of the molecules of a diatomic gas is V. When the temperature is doubled ,the molecules dissociate into two atoms. The new root mean square speed of the atom is : |
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Answer» v According to PROBLEM T will become 2T and M will become `M//2` . So the value of `v_(rms)` will increase by `SQRT4 = 2` TIMES . i.e., NEW root mean square VELOCITY will be 2v. |
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| 37. |
Derive the expression for a body projected horizontally. |
Answer» Solution :(i) At any instant t , the projectile has velocity components along both x - axis and y - axis .  (ii) The resultant of these two components gives the velocity of the projectile at that instant t , as shown in Figure (iii) The velocity component at any t along horizontal (x - axis) is `v_(x)=u_(x)+a_(x)t` SINCE, `u_(x)=u, a_(x)=0,`we get `v_(x)=u.` (iv) The component of velocity along vertical direction (y - axis) is `v_(y)=u_(y)+a_(y)t` Since, `u_(y)=0, a_(y)=G,`we get `v_(y)=g t` (V) Hence the velocity of the particle at any instant is `vecv=uhati+g t hatj` (vi) The speed of the particle at any instant t is GIVEN by `THEREFORE""v=sqrt(v_(x)^(2)+v_(y)^(2)), v=sqrt(u^(2)+g^(2)t^(2))` |
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| 38. |
If the distance between the centres of the atoms of potassium and bromine in KBr (potassium - bromide) molecule is 0.282xx10^(-9)m, find the centre of mass of this particle system from potassium (mass of bromine = 80 u, and of potassium = 39 u). |
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Answer» Solution :Mass of bromine, `m_(Br)=80` units Mass of potassium, `m_(K)=39`units POSITION co-ordinate of potassium, `x_(k)=0` Position co-ordinate of bromine, `x_(Br)=0.282xx10^(-9)m`  `:.` Position co-ordinate of CENTRE of mass, `x_(c)=(m_(k)x_(k)+m_(Br)x_(Br))/(m_(k)+m_(Br))` `x_(c)=(39xx0+80xx0.282xx10^(-9))/(39+80)` `impliesx_(c)=0.189xx10^(-9)m` |
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| 40. |
A mass of lkg attached to one end of a string is first lifted up with acceleration 4.9m//s^2 and then lowered with same acceleration. What is the ratio of tension in string in two cases |
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Answer» 1:3 |
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| 41. |
Maximum ………… error in sum of difference of two quantites is ………… of absolute errors in …………… |
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Answer» |
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| 42. |
In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back. The combined mass of all the persons performing the act, and the tables, plaques etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load. (g = 9.8 ms ^(-2))Young's Modulus for bone Y=9.4xx 10 ^(9) Nm^(-2). |
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Answer» Solution :Tota mass of all the PERFORMERS tables, plaques etc. Is M = 280 kg Mass of performer at the bottom of the PYRAMID `m = 60 kg ` Mass of SUPPORTED by the legs of the performer at the bottom of the pyramid `m _(1) = M - m` `therefore m _(1) =280 -60 = 220kg` Weight of this supported mass, `W = m _(1) G = 220 xx 9.8 =2156N` Weight suppported by each thingbone of the proformer (weight) `= W/2 = (2156)/(2) =1078 N(because` Two thingbone) Length of each thingbone `L = 0.5m` The radius of thingbone `r=2.0 m =2 xx 10 ^(-2) m` Young.s modulus, `Y = (F //A)/(Delta L //L)` `therefore Delta L = (F)/(pi r ^(2)) xx (L)/(Y)` where, `Delta L =` compression in each thingbone. `= (1078)/(3.14 xx (2 xx 10 ^(-2))^(2)) xx (0.5)/(9.4 xx 10 ^(9))=4.565 xx 10 ^(-5) m= 4.56 xx 10 ^(-5)m` The fractional decrease in the thingbone, `(Delta L)/(L) = (4.56xx 10 ^(-5))/(0.5) = 9.12 xx 10 ^(-5) ` `therefore (Delta L )/(L) xx 100 % 9.12 xx 10 ^(-3) %` |
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| 43. |
The sum of all electromagnetic forces between different particles of a system of charged particle is zero |
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Answer» only if all the particles are POSITIVELY CHARGED |
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| 44. |
A ball of mass 2g moving with a vclocity of 2ms^(-1) collides with another ball of mass 8g which is at rests and comes to rest after collission. Thenthe coefficient of restitution is |
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Answer» 1 |
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| 45. |
For the displacement-time graph shown in figure, the ratio of the magnitudes of the speeds during the first two second and the next four second is |
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Answer» `1:1` |
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| 46. |
When we express the velocity of light 30,00,00,000 in standard form up to three significant fogures it is |
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Answer» `3xx10^(8)ms^(-1)` |
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| 47. |
A simple pendulum is suspended from the ceiling of trolley which is sliding down on a inclined plane of inclination theta. Find the angle made by the string with trolley (a) when trolley slides down with uniform velocity (b) when the plane is smooth. |
Answer» Solution :(a) Let T be the tension in the string and the string makes an angle `ALPHA` with the normal to the trolley. In equilibrium, `T cos alpha = Mg cos theta` `T sin alpha = Mg sin theta` `TAN alpha = Tan theta rArr alpha = theta` (b) When trolley is SLIDING on smooth INCLINED plane, the acceleration trolley `a = g sin theta`. The bob of the pendulum experiences a PSEUDO force .ma. opposite to the acceleration of the trolley.  In equilibrium, `T sin alpha + Ma = Mg sin theta` `T sin alpha = Mg sin theta - Ma = Mg sin theta - Mg sin theta` `T sin alpha = 0 rArr sin alpha = 0 rArr alpha = 0` |
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| 48. |
Two liquids of masses M_(1), and M_(2) and specific heats S_(1), and S_(2), respectively are mixed. The specific heat of the mixtureis |
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Answer» `(M_(1)S_(1)+M_(2)S_(2))/(M_(1)+M_(2))` |
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| 49. |
Who among the following first gave the experimental velocity of G? |
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Answer» Cavendish |
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| 50. |
The total kinetic energy of a body of mass 10 kg and radius 0.5 m moving with velocity of 2 m/s without slipping is 32.8 J. The radius of gyration of a body is ……….. |
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Answer» `0.25m` `E=(1)/(2)mv^(2)(1+(k^(2))/(R^(2)))" "["where "V_(CM)=v]` `THEREFORE 32.8xx2=10xx(2)^(2) [1+(k^(2))/((0.5)^(2))]` `therefore (65.6)/(10xx4)=1+(k^(2))/(0.25^(2))` `therefore 1.64=1+(k^(2))/(0.25)` `therefore 0.64=(k^(2))/(0.25)` `therefore k^(2)=0.64xx0.25` `therefore k=0.8xx0.5` `therefore k=0.4m` |
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