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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. |
Select odd man out(incorrect) from the following expressions for kinetic energy in pure rolling. |
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Answer» In pure ROLLING K.E. `=(K.E.)_("TRANS")+(K.E.)_("Rot")` |
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| 2. |
Answer the following questions : (a) The angle subtended at the eye by an object is equal to the angle subtended at the eye the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provided angular magnification ? (b) In viewing through a magnifying glass, one usually positions one's eye very close to the lens. Does angular magnification change if the eye is moved back ? ( c) Magnifying power of a simple a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power ? (d) Why must both the objective and eye piece of a compound microscope have a short focal lengths ? (e) when viewing through a compound microscope, our eyes should be positioned not on the eye piece, but a short distance away from it for best veiwing, why ? How much should be that short distance betweenthe eye and eye piece ? |
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Answer» Solution :(a) It is true that angular size of imahe is equal to angular size of the object. By using magnifying glass, we keep the object far more closer to the eye than at `25 cm`, its normal position without use of glass. The closer object has LARGER angular size than the same object at `25 cm`. It is in this sense that angular magnification is achieved. (b) Yes, the angular magnification changes, if the eye is moved back. This is because angle subtended at the eye would be slightly less than the angle subtended at the lens. The EFFECT is negligible when image is at much larger distance. ( c) Theoretically, it is true. However, when we decrease focal length, aberrations both spherical and chromatic become more pronounced. Further, it is difficult to grind lenses of very small focal lengths. (d) Angular magnification of eye piece is `(1 + (d)/(f_e))`. This increases as `f_e` decreases. Further, magnification of objective lens is `(v)/(u)`. As object LIES close to focus of objective lens `u ~~ f_0`. To increase this magnification `(v//f_0), f_0` should be smaller. (e) The image of objective lens in eye piece is called 'eye ring'. All the rays from the object REFRACTED by the objective go through the eye ring. Therefore, ideal position for our eyes for viewing is this eye ring only. When eye is too close to the eye piece, FIELD of view reduces and eyes fo not collect much of the light. The precise location of the eye would depend upon the separation between the obeject and eye piece, and also on focal length of the eye piece. |
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| 3. |
A block of wood of mass 0.5kg is placed on plane making angle 30° with the horizontal. If the coefficient of friction between the surface of contact of the body and the plane is 0.2, what force is required to keep the body sliding down with uniform velocity? |
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Answer» `6.4N` |
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| 4. |
An artificial satellite of the earth releases a packet. If air resistance is neglected, the point where the packet will hit, will be |
| Answer» Answer :D | |
| 5. |
A planet revolves round the sun in an elliptical orbit of minor and major axes x and y respectively. Then the time period of revolution is proportional to |
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Answer» `(X+y)^(3//2)` |
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| 6. |
Two identical point masses are fixed on Y-axis at equal distances from the origin. A small particle starts on the X-axis from x = oo towards the origin. Its acceleration a is taken positive along its direction of motion. Out of the following graphs the one that best represents the variation of a with x-coordinate a particle is |
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Answer»
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| 7. |
A person standing in the centre of a room, 12m high looks into a plane mirror fixed on the wall. Then the minimum length of the plane mirror for him to see the full length image of the wall behing him, is equal to |
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Answer» 12m |
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| 8. |
The amplitude of vibration of a particle is given by a_(m)=(a_(0))/(aw^(2)-bw+c) where a_(0), a , b and c are positive. The condition for a simple resonant frequency |
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Answer» `B^(2)=4AC` |
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| 9. |
A particle is moving with shm in a straight line. When the distance of the particle from the equilibrium position has the value x_1 and x_2the corresponding values of velocities are v_1and v_2show that period is T=2pi[(x_2^2-x_1^2)/(v_1^2-v_2^2)]^(1//2) |
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Answer» Solution :VELOCITY at a DISTANCE `x_1` is `v_1=omegasqrt(a^2-x_1^2)` and `v_2=OMEGA SQRT(a^2-x_2^2)` SQUARING and subtracting `v_1^2-v_2^2=omega^2[a^2-x_1^2-a^2+x_2^2]` `omega^2=(v_1^2-v_2^2)/(x_2^2-x_1^2)` `omega=[(v_1^2-v_2^2)/(x_2^2-x_1^2)]^(1//2)` `(2pi)/T=omega"or " T=(2pi)/omega` `T = 2pi[(x_2^2-x_1^2)/(v_1^2-v_2^2)]^(1//2)` |
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| 10. |
A and B are two satellites revolving round the earth in circular orbits have time periods 8hr and 1 hr respectively . The ratio of their radius of orbits |
| Answer» ANSWER :C | |
| 11. |
A block of mass m is pulled along a horizontal surface by applying a force at an angle theta with the horizontal. If the block travels with a uniform velocity and has a displacement d and the coefficient of friction is mu, then the work done by the applied force is |
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Answer» `(mu mgd)/(cos theta + mu sin theta)`  Because the block moves with a uniform velocity, the resultant force is zero. Resolving F into horizontal components `F cos theta` and vertical component `F sin theta,` we GET `R + F sin theta = mg orR = mg - F sin theta` Also , `f = mu R = mu(mg - F sin theta)` But `f = F cos theta` `:. F cos theta = mu (mg - F sin theta)` or `F (cos theta + mu sin theta) = mumg` or `F = (mu mg)/(cos theta + mu sin theta theta)` work done, `W = FS cos theta, W= (mu m g d cos theta)/(cos theta + mu sin theta) ( :. s = d)` |
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| 12. |
How are we able to break a wire by repeated bending ? |
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Answer» It is an easy process |
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| 13. |
What is 'Force constant' of a spring? On what factors does it depend? |
| Answer» SOLUTION :Force constant (k) of a spring is defined as the ratio of force APPLIED to displacement PRODUCED in the spring. Force constant `k= F/y`. Force constant k depends on the elastic properties of spring MATERIAL. | |
| 14. |
A constant force F is applied at the top of a ring of that M and radius is R as shown. The angular momentum of particle about point constact at time |
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Answer» is CONSTANT |
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| 15. |
A thin uniform rod of mass m moves translationally with acceleration a due to two antiparallel force of lever arm l. One force is of magnitude F and acts at one extreme end. The length of the rod is (x(F+"ma")l)/("ma"). Where 'x' is |
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| 16. |
Arrange the following simple pendulum in ascending order of their periods of oscillation a.of length 1 m at a place of g= 8m//s b.of length 2m at a place of g=10m//s c. of length 4m at a place of g= 4m//s |
| Answer» Answer :A | |
| 17. |
A mass attached to a spring is free to oscillate, with angular velocity omega, in a horizontal plane without friction or damping. It is pulled to a distance x_0 and pushed towards the centre with a velocity v_0 at time t=0. Determine the amplitude of the resulting oscillaters omega, x_(0)" and "v_(0). [Hint : Start with the equation x= a cos (omega t+theta) andnote that the initial velocity is negative.] |
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Answer» Solution :Displacement at time t in SHM, `x = a cos (omega t+phi)"""…….."(1)` Where `phi` = initial phase, `omega` = angular FREQUENCY and a= amplitude DIFFERENTIATING equation (1) w.r.t. .t., `V= (dx)/(dt)` `=(d)/(dt) [a cos (omega t+phi)]` `therefore v = -a omega sin (omega t+ phi)""".........."(2)` When `t=0, x =x_(0)" and "(dx)/(dt)= -v_(0)` From equation (1), `x_(0) = a cos (omega xx 0+phi)` `x_(0) = a cos phi ""........."(3)` and from equation (2), `-v_(0) = a omega sin (0+phi)""[t=0]` `therefore v_(0) = a omega sin phi` `therefore (v_0)/(omega) = a sin phi"""........."(4)` Squaring and ADDING equation (3) and (4), `x_(0)^(2)+(v_(0)^(2))/(omega^2)= a^(2)cos^(2) phi+ a^(2) sin^(2) phi` `=a^(2) (cos^(2) phi + sin^(2) phi)` `therefore x_(0)^(2) +(v_(0)^(2))/(omega^2) = a^(2)""[therefore cos^(2) phi+ sin^(2) phi= 1]` `therefore a= sqrt(x_(0)^(2) +(v_(0)^(2))/(omega^2))`. |
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| 18. |
Assertion: The difference of two vectors A and B can be treated as the sum of two vectors. Subtraction of vectors can be defined in terms of addition of vectors. |
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| 19. |
Three liquids A, B and C having same specific heat and mass , 2m and 3m have temperatures 20^(@)C, 40^(@)C and 60^(@)C respectively. Temperature of the mixture when (MPP_PHY_C13_E01_274_Q01.png" width="80%"> |
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Answer» |
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| 20. |
Whichof the following pairs is a correctpairA particle revolves round a circular path the acceleration of the particle is |
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Answer» along the circumference of the circle and `a=(V^(2))/(m)` |
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| 21. |
Prove the law of conservation of linear momentum use it to find the recoil velocity ofa gun when a bullet is fired from it |
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Answer» Solution :(i) The forceon eachparticle ( Newton'ssecondlaw) can be written as ` vecF_(12) = ( dvecp_(1))/(dt)and vecF_(21) = ( dvecP_(2))/(dt)` (ii)Here ` vecP_(1)` is the MOMENTUMOF particle1 whichchanges due to the FORCE` vecF_(12)`exertedby particle 2. Further ` vecP_(2)`is the momentumof particle 2. Thischanges due to ` vecF_(21)`exertedby particle 1 . ` ( dvecp_(1))/(dt) = - (dvecp_(2))/( dt)` ` ( dvecP_(1))/( dt)+ ( d vecP_(2))/(dt) = 0` ` d/( dt)( vecP_(1) + vecP_(2)) = 0` (iii) It impliesthat ` vecP_(1) + vecP_(2)` = constantvector ( always) (iv) `vecP_(1) + vecP_(2) ` is the totallinearmomentumof thetwoparticles` ( vecP_(" tot"= vecP_(1) + vecP_(2))` . It is also called as total linear momentumof the SYSTEM. Here , the two particles constitutethe system. (v) Ifthere are no externalforcesactingon thesystem,then the totallinear momentumof thesystem ` (vecP_(" tot"))`is alwaysa constantvector . |
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| 22. |
A body executing S.H.M has maximum acceleration equal to 24m//"sec" and maximum velocity equal to 16m//"sec". The amplitude of S.H.M is |
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Answer» `(3)/(32)m` |
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| 23. |
A particle falls towards the earth from infinity. The velocity with which it reaches earth's surface (g to acceleration due to gravity on earth's surface and R is the radius of the earth) |
| Answer» Answer :B | |
| 24. |
A ball of mass m moving with a speed v makes a head on collision with an identical ball at rest. The kinetic energy after collision of the balls is three fourth of the original kinetic energy. The coefficient of restitution is |
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Answer» `1/2`  Applying the principle of conservation of linear momentum , we get `mv = mv_1 + mv_2 " or " v = v_1 + v_2 "" ….(i)` By DEFINATION of coefficient of restitution `e = (v_2 - v_1)/(u_1 - u_2) = (v_2 - v_1)/(v - 0)" or " v_2 - v_1 = ev "" .....(II)` As KE after COLLISION = `3/4 KE` before collision `:. 1/2 m (v_1^2 + v_2^2) = 3/4 xx 1/2 mv^2 " or " v_1^2 + v_2^2 = 3/4 v^2 "".....(iii)` Squaring eq. (i), we get `v_1^2 + v_2^2 + 2v_1 v_2 = v^2 "".........(IV)` Subtracting (iiii), from EQUATION (iv), we get `2v_1 v_2 = 1/4 v^2"".....(v)` Squaring (ii), we get `v_2^2 + v_1^2 - 2v_1 v_2 = e^2 v^2 "" ....(vi)` Using (iii) and (v) in (vi), we get `3/4 v^2 - 1/4 v^2 = e^2 v^2 , 1/2 v^2 = e^2 v^2 or e = 1/(sqrt2)`. |
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| 25. |
A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plots shown, indicae the one the represents the velocity (v) of the pebble as a function of time (t). |
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Answer» |
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| 26. |
Select the correct alternative: In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies. |
| Answer» SOLUTION :total liner MOMENTUM and total ENERGY of the system of two BODIES . | |
| 27. |
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 kmh^(-1).Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^(-1). What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !] |
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Answer» Solution :Average velocity `= ("Total displacement")/("Time interval")` and average speed `= ("Path length")/("Time interval")` (i) For 0 to 30 minutes time interval, Distance covered from home to MARKET , `=2.5 km` Speed =5 km/hr `therefore` Time taken in market `= (2.5)/(5) = 0.5hours = 30` minutes (a) Average velocity `= ("displacement")/("time") = (2.5)/(0.5) =6km//hr` (B) Average speed `= (2.5)/(0.5) = 5 km//hr` (ii) For 0 to 50 minutes time interval, Velcity while returning back `=7.5 km//hr` distance `=2.5 km` `therefore ` Time taken to cover this distance `= (2.5)/(7.5) = 1/3` HOURS = 20 minutes Time taken to complete the travelling `=30+20 ` (home to market and market to home) = 50 minutes`=5/6` hours Displacement `=2.5 -2.5 =0 km` `therefore` Average velocity `= (0)/(5//6)=0` Total distance covered `=2.5 + 2.5 =5 km` Average speed `= (5)/(5//6) =6` km/hr (iii) For time interval 0 to 40 minutes, Time taken by man to reach market from home, =30 minutes `therefore ` Time to return back, `= 40 -30= 10` minutes Distance covered in 10 minuutes, = velocity while returning `xx` time `=7.5 xx (10)/(60) = 1.25 km` Displacement `=2.5 - 1.25 =1.25 km` `therefore ` Average velocity `= (1.25)/(40//60)=1.875km//hr` Avereage speed `= ("total distance")/("time")` `=(2.5 +1.25)/(40//60) =3.75 xx (60)/(40)=5.625` km/hr In second case, average velocity is zero but not average speed. |
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| 28. |
Three spherical balls of masses 1kg, 2kg and 3kg are placed at the corners of an equilateral triangle of side 1m. The magnitude of the gravitational force exterted by 2 kg and 3kg masses on 1 kg mass. |
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Answer» 19 G |
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| 29. |
A body of mass 5kg is thrown up vertically with a kinetic energy of 1000J. If acceleration due to gravity is 10ms^(-2), find the height at which the kinetic energy becomes half of the original value. |
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Answer» 10m then `PE =1/2 KE` ` MGH = 1000/2 = 500` ` H = (500)/(mg) = (500)/(5 xx 10) = 10m ` |
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| 30. |
Find the maximum speed at which a car can turn round a curve of 36 m radius on a level road. Given the coefficient of friction between the tyre and the road is 0.53. |
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Answer» Solution :Radius of the CURVE r= 36 m Coefficient of FRICTION `mu`= 0.53 Acceleration due to gravity `g= 10 ms^(-2)` `v_(MAX) = sqrt(mu RG) = sqrt( 0.53 xx 36 xx 10 ) = 13.81 ms^(-1)` |
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| 31. |
A force F is given by F at + bt^(2) where t is time. What are dimensions of 'a' and 'b'? |
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Answer» `MLT^(-3)` and `ML^2T^4` |
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| 32. |
It is a well known fact that during a toal solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather fro examples 2.3and 2.4, determine the approximate diameter of the moon. |
| Answer» SOLUTION :3,581 KM | |
| 33. |
A motor is rotating at a constant angular velocity of 600 rpm. The angular displacement in 2 second is ………….. |
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Answer» `40pi` rad `THEREFORE A=omegat=20pixx2=40pi" rad"` |
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| 34. |
K is the force constant of a spring. The work done is increasing its extension from l_(1), to l_(2), will be |
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Answer» `K(l_(2)-l_(1))` |
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| 35. |
Define frame of reference and give its types. |
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Answer» Solution :FRAME of reference : The place or POSITION from which OBSERVER observes the situation (motion) is called from of reference. It has TWO TYPES. (1) Inertial (frame to Reference (2) Non-inertial (frame to) Reference |
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| 36. |
The gravitational force is always attractive in nature. Why? |
| Answer» Solution :Because of negative sign in equation `VEC(F) = -(GM_(1)M_(2))/(r^2) hat^(r)` that the gravitational force is always attractive in nature and the direction of the force is along the LINE JOINING the TWO masses. | |
| 37. |
Which of the following statements are ture/ false ? (i) Keplar discovered famous theory of relativity. (ii) Nuclear reactors are based on controlled nuclear chain reaction. (iii) Einstein explained photoemission on the basis of Planck's quantum theory. |
| Answer» SOLUTION :(i) FALSE (II) TRUE (III) True | |
| 38. |
A satellite is moving in a circular orbit around the earth with a speed equal to half the escape speed from the earth. If 'R' is the radius of the earth, then the height of the satellite above the surface of the earth is |
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Answer» R/2 |
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| 39. |
The lower end of a capillary tube ofdiameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker . What is the pressure required in the tube in order to blow a hemispherical bubble at the end in water ?The surface tension of water at temperature of the experiments is 7.30xx10^(-2)Nm^(9-1). 1 atmospheric pressure =1.01xx10^(5)Pa.Density of water =1000kg//m^(3),g=9.80ms^(-2).Also calulate the excess pressure . |
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Answer» Solution :Radius of capillary tube , `r=(2.0)/(2)mm` `=1.0mm=1XX10^(-3)m` Surface tension of water `=7.30xx10^(-2)Nm^(-1)` Atmospheric PRESSURE `P=1.01xx10^(5)Pa` Density of water `rho=1000kgm^(-3)` Acceleration of gravity `g=9.8ms^(-2)` DEPTH of bubble `h=8.0cm=8xx10^(-2)m` The pressure difference between inside and outside the bubble form in liquid, `P_(i)-P_(o)=(2s)/(r)` The radius if bubble is equal to the radius of capillary tube) `thereforeP_(i)=P_(o)+(2s)/(r)` ...(1) But `P_(o)=P_(a)+hrhog` `=1.01xx10^(5)+8xx10^(-2)xx10^(3)xx9.8` `=1.01xx10^(5)+784` `1.01784xx10^(5)Nm^(-2)` From equation (1), `P_(i)=P_(o)+(2s)/(r)` `=1.01784xx10^(5)+(2xx7.3xx10^(-2))/(1xx10^(-3))` `=1.01784xx10^(5)+14.6xx10` `=1.01784+146` `=1.01930` `=1.02xx10^(5)N//m^(2)` The excess pressure in a bubble , `P_(i)-P_(o)=1.01930xx10^(5)-1.01784xx10^(5)` `=0.00146xx10^(5)` `=146Pa` |
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| 40. |
The temperature of 5 moles of a gas at constantvolume is changed from 100^(@)C to 120^(@)C. The change in internal energy is 80J. The total heat capacity of the gas at constant volume will be in joule/kelvin is |
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Answer» 8 |
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| 41. |
A particle executes SHM has amplitude 6cm. Its acceleration at a distance from the mean position 2cm is 8cm//sec^(2). The maximum speed of the particle is |
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Answer» `8cm//"SEC"` |
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| 42. |
Moment of inertia of a circular ring of mass m and radius r about the normal axis passing through its centre is |
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Answer» `(MR^(2))/(4)` |
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| 43. |
For vectors vec(A) = (2i + 3j - 2k), vec(B) = (5i + nj + k) and vec(C ) = (-I + 2j + 3k) to be coplanar, the value of n is |
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Answer» |
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| 44. |
A man of 60 kg gains 1000 cal of heat by eating 5 mangoes. His efficiency is 56%. To what height he can jump by using this energy? |
| Answer» Answer :A | |
| 45. |
The maximum restoring force of a body executing SHM is alpha and total energy is beta obtain it’s the amplitude in terms of beta" and "alpha. |
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Answer» Solution :Maximum restoring force `F= KA` `therefore alpha = kA` Total energy `E= (1)/(2) kA^(2)` `therefore BETA = (1)/(2) kA^(2)` `therefore (beta)/(alpha)= (A)/(2)` `therefore A= (2beta)/(alpha)`. |
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| 46. |
A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released in starts performing simple harmonic motion at angular frequency omega. If the radius of the bottle is 2.5 cm then omega is close |
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Answer» `2.50 rad"/"s` (Weight of FLOATING object)= (Weight of LIQUID displaced by floating object) `therefore mg = m_(0)g` `therefore m= m_(0) implies V. p_(0)= Alp_(0)` `implies l= (m)/(A p_(0))= (m)/(pi R^(2) p_(0))` `therefore l= (310g)/((3.14)(2.5 cm)^(2)(1-(g)/(cm^3))) approx 16 cm approx 0.16 m` Now angular frequency, `omega = sqrt((g)/(l)) = sqrt((9.8)/(0.16))= sqrt(61.25)= 7.826 (rad)/(s)` The nearest value in GIVEN options for above answer is `5 rad"/"s`. `therefore omega approx 5.0 (rad)/(s)`. 
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| 47. |
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel. |
| Answer» SOLUTION :3675 Non each FRONT wheel, 5145 N on each BACK wheel. | |
| 48. |
A closed cubical box is completely filled wilth water and is accelerated horizontally towards right with an acceleration a. The resultant normal force by the water on the top of the box. |
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Answer» PASSES through the centre of the top |
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| 49. |
Describe the measurement of Earth's shadow (umbra) radius during total lunar eclipse. |
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Answer» Solution :(i) It is possible to measure the radius of shawdow of the Earth at the point where the Moon crosses. (ii) When the Moon is inside the umbra shadow, it appears red in COLOUR. As soon as the Moon exits from the umbra shadow, it appears in crescent shape. (III) By finding the apparent radii of the Earth's Umbra shadow and the Moon, the RATIO of the these radii can be CALCULATED. This is shown in figure.  (iv) Schematic diagram of umbra disk radius. The apparent radius of Earth's umbra shadow = `R_(S) = 13.2 cm` The apparent radius of the Moon = `R_(m) = 5.15 cm` (v) The ratio `(R_s)/(R_m) = 2.56` The radius of the Earth's umbra shadow is `R_(s) = 2.56 xx R_(m)`. |
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| 50. |
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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