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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. |
Two blocks of masses m_(1) = 2 kg and m_(2) = 5 kg are moving in the same direction along a frictionless surface with speeds 10 m//s and 3 m//s, respectively, m_(2) being ahead of m_(1). An ideal spring with k = 1120 N//m is attached to the back side of m_(2). Find the maximum compression of the spring when the blocks collide. What are the final velocities of the blocks when they separate? |
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Answer» SOLUTION :Form conservation of momentum `v=(m_(1)v_(1)+m_(2)v_(2))/(m_(1)+m_(2))=5m//s` conservationof energy `1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)=1/2(m_(1)+m_(2))v^(2)+1/2kx^(2)` `implies kx^(2)=(m_(1)v_(1)^(2)+m_(2)v_(2)^(2)-(m_(1)-m_(2))v^(2))/k` `implies x=sqrt(40/1120)=0.25m` when the blocks have equa SPEEDS, spring has maximum commpression. after his instasnt,the spring againexpands, and after sometimes `m_(1)` loses contact with the spring. Let `v_(1)` and `v_(2)` be the velocities of the blocks after they lose contact. `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)u_(2)` and `v_(1)-v_(2)=-1(u_(1)-u_(2))` `[e=1,` because there is no loss in `KE`] Solving for `v_(1)` and `v_(2)` we get `v_(1)=0m//s` and `v_(2)=7m//s` |
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| 2. |
Which of the following sets of forces acting simultaneously on a particle keep it in equilibrium? |
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Answer» 3N. 5N, 10N |
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| 3. |
A parachutist drops freely from an aeroplane for 8s before the parachute opens out. Then he descends with a net retardation of 2 ms^(-2)reaching the ground with a velocity of 6 ms^(-1). The height from which he bails out of the aeroplane is |
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Answer» 1929 m |
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| 4. |
The dimensional formula of coefficient of viscosity is …………… . |
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Answer» `M^0L^(-1)T^(-1)` |
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| 5. |
Vechile stop on applying brakes does this phenomenon violate the principle of conservation of momentum |
| Answer» SOLUTION :No , it doesnotviolatebacaseexternalforceis actingon THESYSTEM | |
| 6. |
If the earth shrinks to half of its radius and mass remains constant, then the weight of an object on earth will become |
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Answer» doubled |
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| 7. |
A test tube weighing 10 g and external diameter 2 cm is floated vertically in water by placing 10 g of mercury at its bottom. The tube is depressed in water a little and then released. Find the time of oscillation. Take g=10 ms^(-1) |
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Answer» Solution :Total mass of test TUBE and mercutry, `m+10+10=20 g=0.02 KG` Area of cross- section of the test tube tube `A= pi R^(2)=(22)/(7)xx((1)/(100))^(2)` Density of water `rho=10kg m^(-3)` Let the tube be depresed in water by a little distance y and then released. Spring factor, `k=(F)/(y)=(A_(y. rho. g))/(y)=A rho g` `=(22)/(7)NM^(-1)` Interia factor, `m=0.02 kg` `:.T = 2pi sqrt((m)/(k))` `=2xx(22)/(7)xx sqrt(0.02xx7)/(22)` `=0.5 ` |
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| 8. |
Figureshows the position-time graph of a body of mass 0.04 kg.Suggest a suitable physical context for this motion.What is the time between two consecutive impulses received by the body ? What is the magnitude of each impulse ? |
| Answer» SOLUTION :A body with a constant speed of `1 cm s^(-1)`receives impulse of MAGNITUDE `0.04 kgxx 0.02 m s^(-1) =8xx 10^(-4) kg m s^(-1)`after every2 s from the WALLS at x = 0 and x = 2 cm. | |
| 9. |
The potential energy of a system increases if work is done |
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Answer» by the SYSTEM against a CONSERVATIVE FORCE |
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| 10. |
What is a sonometer? Give its construction and working. Explain how to determine the frequency of tuning fork using sonometer. |
Answer» Solution :Stationary waves in sonometer: Sono means sound related, and sonometer implies sound-related measurements. It is a device for demonstrating the relationship between the frequency of the sound produced in the transverse standing wave in a string, and the tension, length and mass per unit length of the string. Therefore, using this device, we . can determine the following quantities:  (a) the frequency of the tuning FORK or frequency of alternating current ( b) the tension in the string( c) the unknown hanging mass Construction: The sonometer is MADEUP of a hollow box which is one meter long with a uniform metallic thin string attached to it. One end of the string is connected to a hook and the other end is connected to a weight HANGER through a pulley as SHOWN in figure.Since only one string is used, it is also known as monochord. The weights are added to the free end of the wire to increase the tension of the wire. Two adjustable wooden knives are put over the board, and their positions are adjusted to change the vibrating length of the stretched wire. Working: A transverse stationary or standing wave is produced and hence, at the knife edges p and Q, nodes are formed. In between the knife edges, anti-nodes are formed. If the length of the vibrating element is lthen `l = lamda/2 rArr lamda = 2l` Let f be the frequency of the vibrating element, T the tension of in the string and `mu`the mass per unit length of the string. Then using equation ,we get ` f = v/lamda = (1)/(2l) sqrt(T/mu) `in Hertz....(1) Let `rho`be the density of the material of the string and d be the diameter of the string. Then the mass per unit length `mu` , `mu =" area " xx "density " = pi r^2 rho= (pi rho d^2)/(4)` Frequency ` f = v/lamda = (1)/(2l) sqrt( (T)/((pid^2 rho)/(4) ) )` ` therefore f =(1)/(ld) sqrt( (T)/(pi rho) ) ` |
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| 11. |
What is meant by an open pipe system ? |
| Answer» Solution :A pipe with both ENDS OPEN is KNOWN as an open system. | |
| 12. |
Five moles of hydrogen is heated through 20Kunder constant pressure. If R=8.312J/mole K find the external work done. |
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Answer» 831.2J |
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| 13. |
A metre scale of mass 100 gm is suspended freely at its end. It is pulled aside and swung about a horizontal axis with an angular velocity 5 rad/s. Calculate the height to which the centre of mass of the scale rises above its lowest position, assuming that there is no friction and air resistance. (given : length of meter scale used = 1m ) |
| Answer» Solution :`(1//2) Iomega^(2) =mgh, I=(1//3) ml^(2), h = L^(2)OMEGA^(2), h = l^(2)omega^(2)//6G = 1^(2) xx 5^(2)//6 xx 9.8=0.425 m` | |
| 14. |
In a compound microscope the object is 1cm from the objective lens. The lenses are 30cm apart and the intermediate image is 5cm from the eyepiece. What magnification is produced? |
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Answer» SOLUTION :As the LENSES are 30cm apart and intermediate image is formed 5cm in FRONT of eye lens, `u_e=5cm and v=L-u_e=30-5=25cm` Now as in case of compound microscope `M=m_0 times m_e=-v/u times [D/u_e]` Here u=1cm and D=25cm So, `M=-25/1 times [25/5]=-125` Negative sign impiles that FIND image is inverted |
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| 15. |
Suppose the universe were shrunk to the size of the earth how large would the earth be onn such a scale? |
| Answer» SOLUTION :About the SIZE of the ATOM i.e. `10^(-11)`m | |
| 16. |
In Davisson-Germer experiment the relation between the angle of diffraction theta and the grazing angle 0phi is |
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Answer» `THETA=90^(@)-(PHI)/2` |
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| 17. |
the cylinder of a water pump is 4 cm above the surface of water in a well. If the readius of the cylinder of the pump be 5cm, calculate th e force acting on the piston while the pump is working.(Density of water = 1000 kg m^(-3)) [ hint : force = areax pressure due to 4-m high column of water .] |
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| 18. |
The voltage across a wire is (100pm5)V and the current passing through it is (10pm0.2) A. Find the resistance of the wire. |
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Answer» Solution :VOLTAGE `V = (100 pm 5) V` Current I = (10 pm 0.2) A` Resistance R = ? Then resistance R is GIVEN by Ohm.s law R = V/I = 100/10 = 10 OMEGA ` ` (Delta R)/(R ) = ( (DELTAV)/(V) + (Delta I)/(I) )` `Delta R = ((Delta V)/(V) + (Delta I)/(I) R= (5/100 + (0.2)/(10) ) 10` ` = (0.05 + 0.02)10 = 0.07 xx 10 = 0.7` The resistance ` R = (10 pm 0.7) rho ` |
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| 19. |
When equal volumes of two metals are mixed together, the specific gravity of alloy is 4. When equal masses of the same two metals are mixed together, the specific gravity of the alloy now becomes 3. Find specific gravity of the alloy now becomes 3. Find specific gravity of each metal. |
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Answer» SOLUTION :In CASE of mixture, `rho_("mix")=(m_(1)+m_(2))/(v_(1)+v_(2))`. When equal volumes are mixed, `4=(vrho_(1)+vrho_(2))/(V+v)=(rho_(1)+rho_(2))/(2)......(1)` When equal masses are mixed, `3=(m+m)/((m)/(rho_(1))+(m)/(rho_(2)))=(2rho_(1)rho_(2))/(rho_(1)+rho_(2))....(II)` Therefore, from (i) and (ii) specific gravity of the metals are 2 and 6. |
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| 20. |
A satellite is revolving round the earth in a circular orbit with a velocity of 8km/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 the height, then `therefore a = (V^(2))/(r) = g_(H)rArr (V^(2))/(r) = 8 = (64 xx 10^(6))/(R+h) = 8` `rArr R + h = 8 xx 10^(6)` `h = 8 xx 10^(6) - 6 xx 10^(6) = 2 xx 10^(6)m` = 2000 km |
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| 21. |
A circular platform is mounted on a vertical frictionless axle. Its radius is r=2m and its moment of inertia is I=200kg-m^(2). It is initially at rest. A 70kg man stands on the edge at speed v_(0)=1.0m//s relative to the ground. The angular velocity of the platform is: |
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Answer» `1.2rad//s` |
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| 22. |
A plot of Farenheit temperature (t_(F)) versus celsius temperature |
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Answer»
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| 23. |
Pressure is a scalar quantity because |
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Answer» It is the ratio of FORCE to area and both force and area are vectors |
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| 24. |
When a uniform rod is heated at one of its ends will the position of the centre of gravity be shifted ? |
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Answer» |
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| 25. |
If a disc is given a linear velocity on a rough horizontal surface, then its angular momentum is |
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Answer» CONSERVED about COM only |
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| 26. |
A shell is fired from a long range gun with an initial velocity u=1000ms^(-1) at an angle 30^(@) to the horizon. How long will the shell be in the air? At what distance from the gun will it fall to the ground ? The gun and the pont where the shell lands are on the same horizontal line. |
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Answer» |
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| 27. |
If the size of a nucleus (in the range of 10^(-15)m"to"10^(-14)m) is scaled up to the tip ofa sharp pin , what roughly is the size of an atom ? Assume tip of the pin to be in the range 10^(-5) "to" 10^(-4)m. |
| Answer» Solution :The size of a nucleus is in the range of `10^(-15)m "to" 10^(-14)m`. The tip of a sharp pin I s TAKEN to be in the range of `10^(-15)m "to" 10^(-4)m.` Thus W are scaling up by a factor of `10^(-10)m`. An atom roughly of size `10^(-10)m` will be scaled up to a size of 1m .thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre LONG. | |
| 28. |
A ball with a momentum p falls on a floor vertically and bounces a number of times . If thecoefficient of restitution is e, momentum transferred to the floor is |
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Answer» <P>p(1+e) |
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| 29. |
The acceleration of the sliding object in an inclined plane |
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Answer» `g cos theta ` |
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| 30. |
Two objects of masses 2 kg and 4 kg are moving with the same momentum of 20 kg ms (a) Will they have some kinetic energy! (b) Will they have same speed? |
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Answer» Solution :(a) `"The kinetic energy of the mass is GIVEN by KE"=p^(2)/2M` `" For the OBJECT of mass 2 kg, kinetic energy is KE"_(1)=(20)^(2)/(2xx2)=(4000)/4=100J` `"For the object of mass 4 kg, kinetic energy is KE"_(2)=(20)^(2)/(2xx4)=(4000)/8=50J` Note that `KE_(1)ne= KE_(2)`i.e ., even though both are having the same momentum, the kinetic energy of both masses is not the same. The kinetic energy of the HEAVIER object has lesser kinetic energy than smaller mat It is because the kinetic energy is inversely proportional to the mass `(KEinfty1/m)` for a given momentum (b) As the momentum, p=mv, the two objects will not have same speed, |
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| 31. |
A cannon is fixed with a smooth massive trolley car atan angle theta as shown in the figure. The trolley te trolley car slides from rest down the inclined plane of inclimation beta. The muzzle velocity of the shell fired at t=t_(0) from the cannon is u, such that the shell moves perpendicu lar to the inclined plane just after the firing. The value of r_(0) is: |
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Answer» `(ucos theta)/(G)` |
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| 32. |
A cannon is fixed with a smooth massive trolley car atan angle theta as shown in the figure. The trolley te trolley car slides from rest down the inclined plane of inclimation beta. The muzzle velocity of the shell fired at t=t_(0) from the cannon is u, such that the shell moves perpendicu lar to the inclined plane just after the firing. The difference in range of the shell relative to the trolley car and ground is: |
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Answer» `(U^(2)sin 2 theta)/(G COS beta)` |
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| 33. |
(A) : When a person is walking horizontally with a suitcase on his head, no work is done by him against gravitational force.(R ) : gravitational force on suitcase acts vertically downwards and motion is in horizontal direction, hence dot product becomes zero. |
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Answer» Both (A) and (R ) are true and (R ) is the correct explanation of (A) |
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| 34. |
The force required to stretch a steel wire 1 cm^2 in cross section to double its length is (given Y = 2 xx 10^(11) Nm^(-2)) ………. . |
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Answer» (a) `10^(7) N` `F = 2 xx 10^(11) xx 10^(-4) = 2 xx 10^(7) N` |
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| 35. |
The reading of pressure meter attached with a closed pipe is 5 xx 10^(5) Nm^(-2). On opening the valve of the pipe, the reading of the pressure meter is 4.5 xx 10^(5) Nm^(-2). Calculate the speed of the water flowing in the pipe. |
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Answer» Solution :USING Bernoulli.s equation ` P_1 + 1/2 RHO V_(I )^(2)= P_2+ 1/21/2rho V_(2) ^(2)` ` 1/2 rho( v_(2) ^(2)- V_(1) ^(2)) = P_1 -P_2` Here initial velocity `V_1= 0` and density of water ` rho= 1000 KG m^(-3)` `1/2 xx 10 ^(3)xx V_(2) ^(2)= ( 5-4.5 ) xx 10^(5)` `V_(2)^(2) = ( 0.5 xx 10 ^(5) xx 2 )/( 10 ^(3) = 1 xx 10 ^2` ` V_(2)^(2)= ( 0.5xx 10^(5) xx 2 )/( 10^3) = 1 xx 10^2 ` `V_(2)^(2) = 100` ` V_2 = 10 ms^(-1)` |
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| 36. |
Newton's laws can be applied to mechanical waves, propagating in elastic media. Can they be applied to electromagnetic waves also ? |
| Answer» Solution :No. for ELECTROMAGNETIC WAVES, we need to make some MODIFICATION in Newton.s LAWS. | |
| 37. |
The value of .g. at a particular point on the surface of the earth is 9.8ms^(-2). Suppose the earth suddenly shrinks uniformly to half its present size without losing mass, then the value of .g. on the surface of new earth will be: |
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Answer» `9.8 MS^(-2)` |
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| 38. |
What is the expression between the pressure and kinetic energy of a gas molecule ? |
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Answer» `P=1/2E` |
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| 39. |
In the last question, the axis of the disc is tilted slightly to make an angletheta with the vertical. Redo the problem for this condition and check the result by putting theta = 0 in your answer. |
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Answer» |
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| 40. |
A small object is placed 10cm infront of a plane mirror. If you stand behind the object 30cm from the mirror and look at its image, the distance focused for your eye will be |
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Answer» 60cm |
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| 41. |
A physical quantity P is related to four observables a, b, c and d as follows: P = (a^(3)b^(2))/(dsqrt(c)) The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage errors in the quantity P? |
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Answer» Solution :RELATIVE error in P is GIVEN by `(DeltaP)/(P) = 3 (DELTAA)/(a) +2 (Deltab)/(b) + 1/2 (Deltac)/(c) + (Deltad)/(d)` `(DeltaP)/(P) xx 100 = 3((Deltaa)/(a) xx 100) +2 ((Deltab)/(b) xx 100) +1/2 ((Deltac)/(c) xx 100)+ (Deltad)/(d) xx 100` = `(3xx1%) + (2xx3%)+((1)/(2)xx4%)+(1xx2%) =13%` ROUNDED off value of P = 3.8 |
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| 42. |
A ball of mass 3M rolls from rest on a smooth flat table towards another ball of mass 2M at rest with a constant velocity 5m/s. When the two balls are separated by a distance 6m, the position of centre of mass from the mass 3M is |
| Answer» Answer :B | |
| 43. |
A paralle beam of white light falls on a convex lens. Images of blue, yellow and red light are formed on other side of the lens at a distance of 0.20m, 0.205m and 0.214m respectively. The dispersive power of the material of the lens will be |
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Answer» `619/1000` |
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| 44. |
A body is projected with an initial velocity of 58.8 m/s at angle 60° with the vertical. The vertical component of velocity after 2 sec is |
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Answer» 9.8 m/s upwards |
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| 45. |
In the above given problem if the lower thread is pulled with a jerk, what happens |
| Answer» SOLUTION :When the lowerthreadCDis pulledwith a jerkthe threadCDitselfbreak.Beacausepull on threadCD is nottransmittedto thethreadABinstantly. | |
| 46. |
The acceleration of a particle in ms^(-1) is given by a=3t^(2)+2t+2 where timer is in second If the particle starts with a velocity v=2ms^(-1) at t=0 then find the velocity at the end of 2s. |
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Answer» SOLUTION :`a=(dv)/(dt)=(3T^(2)+2t+2)dt` `dv=(3t+2t+2)dt` `intdv=int(3t^(2)+2t+2)dt` `v=t^(3)+t^(2)+2t+C` `c=2m//s,v=18m//s` at t=2s |
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| 47. |
Imagine a light planet revolving around a very massive star in a circular orbit of radius .r. with a period of revolution T. On what power of r will the square of time period depend if the gravitational force of attraction between the planet and the star is proportional to r^(-5//2) |
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Answer» Solution :Here GRAVITATIONAL force PROVIDES centripetal force, `(mv^2)/( r) = (K)/(r^(5//2)) rArr V^2 = (K)/(MR^(3//2)) rArr T = (2pi r)/(v) = 2pi r sqrt((mr^(3//2))/(K)) rArr T^2 prop r^(7//2)` |
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| 48. |
The mass of a uniform ladder of length 6m is 20kg. A mass 60kg is placed on the ladder at a height of 2m from the bottom. The position of centre of mass from the bottom is |
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Answer» `1.5m` |
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| 49. |
The ratio of the speeds of the objects of masses m_(1), m_(2), and my respectively are fallen freely from a same point 'O' when reaches at ground is ...... |
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Answer» `m_(1) : m_(2):m_(3)` `v ^(2) - v _(0)^(2) =2ad` ` v ^(2) = 2 gH (because v _(0) = 0,a = G, d = H)` `therefore v ` is INDEPENDENT of mass. `therefore v ^(2) ` is constant. `therefore` Ratio` =1:1:1` |
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| 50. |
The wavelength of two sine waves are lambda_(1) = 1 m and lambda_(2) = 6 m . Calculate the corresponding wave numbers . |
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Answer» `1.05radm^(-1)and 6.28rad m^(-1)` ` K_1= ( 2pi )/( lamda_1) = ( 2pi )/( 1) = 6.28rad m^(-1)` ` K_2 =(2pi )/( lamda_2) =(2pi )/( 6 )=1.05rad m^(-1)` |
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