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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. |
One side of an iron plate one metre square is kept at 100°C and the other side is at 0°C. The thickness of the plate is 1 cm. If all the heat conducted across the plate in one minute is used in melting ice, calculate the amount of ice so melted. Thermal conductivity of iron = 0.162 in C.G.S. unit L.H. of fusion of ice = 80 cal/gm. |
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Answer» |
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| 2. |
(A): Heavy boxes are loaded along with empty boxes on a cart. Heavy boxes should be kept at the centre of gravity of the cart, for more stability (R) : The vertical line passing through the centre of gravity should fall within the base of cart for stability, |
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Answer» Both .A. and .R. are TRUE and .R. is the CORRECT EXPLANATION of .A. |
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| 3. |
The mass of the earth is 6xx10^(24) kg and that of the Moon is7.4 xx10^(22) kg. The constant of gravitation G is 6.67 xx10^(-11) "Nm"^(2) "kg"^(-2) The potential energy of the system is-7.79xx10^(28)J The mean distance between the Earth and Moon is .......... metre. |
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Answer» `3.80 xx10^(8)` |
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| 4. |
The apparent weight of a man in a lift is w_1 when lift moves upwards with some acceleration and is w_2 when it is accelerating down with same acceleration. Find the true weight of the man and acceleration of lift. |
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Answer» ` (w_1+w_2)/2` |
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| 5. |
A ball thrown horizontally with velocity v from the top of a tower of heigh h reaches the ground in t seconds. If another ball of double the mass is thrown horizontally with velocity 3v from the top of another lower of height 4h it reaches the ground in (seconds) |
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Answer» `3t//4` |
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| 6. |
Example thermal conductivity and coeifficient of the thermal conductivity. A copper bar of the thermal conductivity 401 W (mk) has one end at 104^(@)C. The length of the bar is 0.10 m and the cross-sectional area is 1.0xx10^(-6) m ^(-2). What is the rate of heat conduction along the bar ? |
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Answer» SOLUTION :The ability to conduct HEAT in solids is called " THERMAL conductivity ". Consider a bar with rectangular crossection as show in the figure . The faces ABCD and EFGH are maintained at `theta_(1) and theta _(2)` respectively `(theta_(0) gt theta_(2))`. Heat passes from one end to the other.  The amount of heat conducted (Q) depends on, (1) Amount of heat conducted Q is proportional to area of cross -section A prependicular to flow. `therefore Qprop A "" ........................(1)` (2) Is proportional to temperature diffrence between the two ends `therefore Qprop (theta_(2)-theta_(1))"" ........................(2)` (3) Is proportional to the time of flow. `therefore Qprop t"" ........................(3)` and (4) Is inversely proportional to the length of the conductor. `therefore Qprop (1)/(l)"" ........................(4)` From the above laws `Q prop (A(theta_(2)- theta _(1))t)/(l) or Q=(KA(theta _(2) -theta _(1))t)/(l) ` where K= constant called coefficient of thermal conductivity . Coefficient of thermal conductivity (K): It is defined as the amount of heat conducted normally per sec per unit area of cross-section per unit temperature gradient. S.I. Unit w m `K^(-1)` Dimensional formula`=[M^(1)L^(1)T^(-3) theta ^(-1)]` Problem: Thermal conductivity of copper , `K_(c)=401 W//m-K ` Temperatureat one end , `theta _(2)=104^(@) C` Temperature of 2nd end, `theta _(1)=24^(@)C` Length of copper bar, `l=0.1m, ` Area, `A=1.0xx10^(-6)m^(-2)` Rate of conduction. `(Q)/(t)=(KA(theta _(2)- theta _(1)))/(l) =(401xx1xx10^( -6)(104-24))/(0.1)=401xx80xx10^(-5)=0.3208 J//S` |
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| 7. |
What is the influence of normal reaction on the area of contact? |
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Answer» SOLUTION :Area of CONTACT INCREASES with NORMAL reaction. |
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| 8. |
In a Millikan's oil drop experiment what is the terminal speed of a drop of radius 2.0 xx 10^(-5)m, and density 1.2 xx 10^(3) kg m^(-3) ?. Take the viscosity of air at the temperature of the experiment to be 1.8 xx 10^(-5) Ns m^(-2)How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop in air. |
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| 9. |
A body is in pure rotation. The linear speed v of the particle, the distance r of the particle from the axis and the angular velocity omega of the body are related as omega=v/r. Thus |
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Answer» `omegaalpha1/R` |
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| 10. |
Dust particles in the suspended state in a mono atomic gas are in thermal aquilibrium with the gas.If the temperature of the gas is 300K,find the mean KE of translation of dust particle,If the mass of a certain particle is 10^(-17) kg,calculate its rms speed (k=1.38xx10^(-23)J/K). |
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Answer» Solution :According to kinetic theory of gases,KE=(3/2)kT, so for a particle in thermal equilibrium `T_(P)=T_(G)=T` so `(KE)_(PA)=(3)/(2)kT=(3)/(2)xx1.38xx10^(-23)xx300=6.21xx10^(-21)J` Now if .m, is the mass of particle and `v_(rms)` its rms SPEED, KE=`(1)/(2)mv_(rms)^(2)`, i.e, `V_(rms)=sqrt((2KE)/(m))` `therefore v_(rms)=sqrt((2xx6.21xx10^(-21))/(10^(-178)))=3.524xx10^(-2)m//s` |
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| 11. |
The speed of a body moving in uniform circular motion is doubled and the radius of the circular path is halved. What happens to the centripetal force? |
| Answer» SOLUTION :CENTRIPETAL force `=(mv^(2))/R` when v becoems 2V and r BECOMES `r//2`, the centripetal force `=(m(2v)^(2))/(r//2) = (mv^(2))/r xx 8`. The force, becomes 8 times its oroginal volume. | |
| 12. |
(A) : If earth shrink to half its present size, length of the day would become 6 hours.(R ) : As size of the earth changes is moment of inertia changes. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 13. |
A stone is thrown horizontally from a height with a velocity v_9x)=15m//s. Determine the normal and tangetial acceleartion of the stone in 1 second after begins to move. |
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Answer» Solution :The HORIZONTAL component of acceleration is zero. The net acceleration of the STONE is directed vertically downward and is equal to the acceleration due to gravity g. Thus `a= g = sqrt(a_(1)^(2)+a_(N)^(2))` from figure we can see that `costheta = (V_("X"))/(V) = (a_( c))/(a) = (a_(g))/(g)` and `sintheta = (V_(y))/(v) = (a_(t))/(a) = (a_(t))/(g)` Hence `a_(t) = g(V_(y))/(V) = (g^(2)t)/(sqrt(v_(x)^(2)+g^(2)t^(2))` and ` a_( c) = g"(v_(x))/(v) = (gv_(x))/(sqrt(v_(x)^(2)+g^(2)t^(2))` on substituting numerical values, `v_(x) = 15m//s, g = 9.8m//s^(2)`, we get `a_(t)=5.4m//s^(2) and a_(n) = 8.2m//s^(2)` |
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| 14. |
A particle is executing simple harmonic motion with an amplitude A and time period T. The displacement of the particles after 2T period from its initial position is |
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Answer» A |
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| 15. |
Assertion: In spring ball model. displace any ball form the its |
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| 16. |
A body of mass m is made to revolve along a circle of radius r in horizontal plane at a speed Vm//s with help of light horizontal string such that the tension in the string is 10 N. Keeping speed constant if mass is increased to 2m und radius is decreased to r//2 then tension in the string is |
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Answer» 10N |
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| 17. |
For a certain gas the ratio of specific heats is given to be gamma=1.5 For this gas ………….. . |
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Answer» <P>`C_(V)=3R//J` `"Now,"C_(P)-C_(V)=RrArr C_(P)-(2)/(3)C_(P)=R rArr C_(P)=3R` |
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| 18. |
Two bars are unstressed and have lengths of 25 cm and 30 cm at 20^(@)C as shown in Figure. Bar(1) is of aluminium and bar(2) is of steel. The cross-sectional areas of the bars are 20 cm^(2) for aluminium and 10 cm^(2) for steel. Assuming that the top and bottom supports are rigid, compute the stress in each member when the temperature is 70^(@) C. Take Y_(a) = 0.70 xx 10^(5)N//mm^(2). Y_(s) = 2.1 xx 10^(5)N//mm^(2). alpha_(a) = 24 xx 10^(-6) //^(@) C and alpha_(2) = 12 xx 10^(-6) //^(@) C |
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Answer» Solution :Given Data : Length of ALUMINIUM, `" " L_(a) = 25 ` cm = 250 mm Length of steel , `"" L_(s) ` = 30 cm = 300 mm Cross - sectional area of aluminium `A_(s)= 20 cm^(2) = 2000 mm^(2)` . Cross - sectional area of steel, `A_(s) = 10 cm^(2) = 1000 mm^(2) . Y_(a) = 0.7 xx 10^(5) N//mm^(2)` `Y_(s) = 2.1 xx 10^(5) N//mm^(2)` `alpha_(a) = 24 xx 10^(-6) //^(@)C and alpha_(s) = 12 xx 10^(-6) //^(@) C ` Due to rise in temperature, both bars will tend to increase in length. As the SUPPORTS are rigid, both of them will be under compression, but the total length will remain unchanged. HENCE, contraction of the two bars due to compressive stress = ELONGATION of the two bars due to rise in temperature Let `simga_(s) = ` compressive stress in steel `simga_(a)`= compressive stress in aluminium T = Rise in temperature = 70 - 20 `= 50^(@)` C Contraction of the two bars due to compressive stress = `(sigma_(s))/(2.1 xx 10^(5)) xx 300 + (sigma_(a))/(0.7 xx 10^(5)) xx 250 `..... (1) Elongation of the two bars to rise of temperature `alpha_(s) . T. L_(s) + alpha_(a). T. L_(a).` `= 12 xx 10^(-6) xx 50 xx 300 + 24 xx 10^(-6) xx 50 xx 250` = `10^(-6) [ 48 xx 10^(4)] ` = 0.48 mm .......... (2) Equating equations (1) and (2) , we get `(sigma_(s))/(2.1 xx 10^(5)) xx 300 + (sigma_(a))/(0.7 xx 10^(5)) xx 250` `142.86 S_(s) + 357 S_(a) = `48000 Also, force in steel = force in aluminium `singa_(s) xx A_(s)""sigma_(a) xx A_(a) " " sigma_(s) xx 1000 = sigma_(s) xx 2000` `sigma_(s) = 2 sigma_(a) `.......... (3) From equation (3) substituting the value of `sigma_(s) = 2 sigma_(a),` we get `142.86 xx 2 sigma_(a) + 357 sigma_(a) = 48,000` `alpha_(a) ` = 74.68 N/`mm^(2)` Ans. `sigma_(s)` = 149.36 N/`mm^(2)` Ans. |
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| 19. |
The horizontal and vertical displacement of a projecticleare given by x=12t and y=16t-5t^(2),all the quantities being measured in S.I. system. The maximum height of the projectile is (g=10ms^(-2)) |
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Answer» 25.6 m |
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| 20. |
Inside a uniform spherical shell (a) the gravitational potential is zero (b) The gravitational field is zero (c) the gravitational potential is same everywhere (d) the gravitational field is same everywhere |
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Answer» only a & B are TRUE |
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| 21. |
The displacementequation of a particle executing simple harmonic motion is x= (5m) sin ( pi t + pi/2) . Find amplitude , time period and maximum speed ? |
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Answer» Solution :Comparing with the EQUATION x=`A sin (omega t + PHI)` , we have amplitude = 5M Time PERIOD `=(2pi)/(omega) = (2pi)/(pi) = 2s` , maximum speed `=A omega =5 xx pi= 5 pi` m/s |
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| 22. |
What is centripetal acceleration or radial acceleration or normal acceleration ? |
| Answer» Solution :In the case of VELOCITY is tangential at EVERY point in the circle , the ACCELERATION is acting towards the centre of the circle . This is CALLED centripetal acceleration . It ALWAYS points towards the centre of the circle | |
| 23. |
A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as alpha=ksqrt(v), where k is a positive constant. At the initial moment the velocity of the point is equal to V_(0). What distance will it take to cover that distance? |
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Answer» Solution :Let `t_(0)` be the TIME in which it COMES to a stop. Given that `-(dv)/(dt)=ksqrt(V)` `underset(0)overset(t_(0))intkdt=underset(v_(0))overset(0)int-(dv)/(sqrt(v))` `because kt_(0)=2sqrt(v_(0))` `therefore t_(0)=(2)/(k)sqrt(v_(0))` Now to find the distance covered before STOPPING, `(dv)/(dt)=(dv)/(ds)(ds)/(dt)=v(dv)/(ds)" But, "(dv)/(dt)=-ksqrt(V),` `therefore v(dv)/(ds)=-ksqrt(V) therefore sqrt(v)dv=-kds` `therefore underset(V_(0))overset(0)intsqrt(v)dv=-underset(0)overset(s)intkdsimpliess=(2)/(3k)V_(0)^((3)/(2))` |
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| 24. |
Vertices of a triangle are given by hati+3hatj+2hatk , 2hati-hatj+hatk and -hati+2hatj+3hatk then area of triangle is (in units) |
| Answer» Answer :A | |
| 25. |
What is the ratio of the angular speeds of the minute hand second hand of a clock ? |
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Answer» `1 : 12` Angular speed of second hand`omega_(s) = (1)/(3600)` ` omega_(m) : omega_(s) = (1)/(60): (1)/(3600)` `= (1)/(1) : (1)/(60)` = 60 :1 |
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| 26. |
A 10sqrt(3) kg box has to move up an inclined slope of 60^(@) to the horizontal at a uniform velocity of 5 ms^(-1). If the frictional force retarding the motion is 150N, the minimum force applied parallel to inclined plane to move up is (g=10 ms^(-2)) |
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Answer» `300xx(2)/(sqrt(3))N` |
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| 27. |
Babies fall down frequently while trying to walk. Why? |
| Answer» SOLUTION :The centre of gravity of a baby, while crawling, remains quitelow and also the line of action of its body weight always passes through the base. As the baby TRIES to stand the centre of gravity of the body GOES up and so the body becomes unstable. To keep the body stable the baby should keep its body erect so that the line of action of the body weight passes through the base. HENCE due to lack of SUFFICIENT practice, babies fall down frequently. | |
| 28. |
Show that the resultant of two vectors vec(A) and vec(B)always lies between A + B andA - B |
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Answer» Solution :Let `theta ` bethe ANGLE between `VEC(A) and vec(B)`. Thus the resultant of `vec(A) and vec(B)`is given by `vec(R) = sqrt(A^(2) + B^(2) = 2AB cos theta)` (or)`R^(2) = A^(2) + B^(2) + 2AB cos theta` Now greatest value of cos`theta `is + 1 Greatest value of `R^(2) "is" (A^(2) +B^(2) +2AB) or(A+ B)^(2)` i.e., the greatest valure of R is (A + B) The least of `cos theta` is - 1 So the least value of `R^(2) is(A^(2) +B^(2) - 2AB) or (A - B)^(2)`i.e., the least value of R is (A - B) |
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| 29. |
The escape velocities on the surface of two planets of masses m_1 and m_2 and having the radius r_1 and r_2arev_1 and v_2respectively, then |
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Answer» `v_1/v_2=(m_1r_1)/(m_2r_2)` |
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| 30. |
Negative gradient of potential energy gives |
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Answer» CONSERVATIVE force |
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| 31. |
Does the change in gravitational potential energy of a body between to given points depend upon the nature of path followed why? |
| Answer» Solution :The change in gravitational Potential energy of a body depends only upon the POSITION of the PATH FOLLOWED. It is due to the fact that the gravitational FORCE is a conservative force depends only on the position of initial & final POINTS and is independent of path followed. | |
| 32. |
Prove that second derivative of position vector is not zero then there must be a force on the body |
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Answer» Solution :The accelerationis the secondderivative of positionvectorof the body`(VEC(a) = (d^(2)vec(R ))/(dt^(2)))` theforceon THEBODY is If`(d^(2) r )/( dt^(2))ne0, vec(F ) =ne 0` |
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| 33. |
Resistance of a given wire is obtained by mea- suring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each then error in the value of resistance of the wire is : |
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Answer» 0 |
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| 34. |
A body is projected with velocity 24ms^(-1) making an angle 30^(@) with the horizontal. The vertical component of its velocity after 2s is (g=10ms^(-1)) |
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Answer» `8 MS^(-1)` UPWARD |
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| 35. |
A metal wire of uniform cross-sectional area A and length L, has mass m. It is suspended vertically from a ceiling. If its Young's modulus is Y, then the elongation Deltal of wire due to its own weight will be ......... |
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Answer» zero Hence the tension at this point = WEIGHT of wire below this point `thereforeT = ((L -x )W)/(L)` The elongation in this element `= ("original length" XX "stress")/(Y)` `= (Tdx)/(AY) = ((L -x ) Edx)/(LAY)` `THEREFORE ` Total elongation in length `= int _(0) ^(L) ((L -x) W dx )/( LAY) = (W)/( LAY) int _(0) ^(L) (L-x) dx` `= (W)/( LAY) [ Lx - (x ^(2))/(2) ] _(0) ^(L) = (WL )/( 2 AY) = (mgL)/(2 AY)` |
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| 36. |
A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be: |
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Answer» 225 |
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| 37. |
A square hole is punched out of a circular lemena, the diagonal of the square coinciding with the radius of circle. If .a. be the diameter of the circle, the distance of the centre of mass of reminder from previous centre of mass. |
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Answer» `(4(2pi-1))/(a)` |
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| 38. |
A body executes SHM, such that its velocity at the mean position 1ms^(-1) and acceleration at the extreme position is 1.57ms^(-2). Calculate the amplitude and the time period of oscillation. |
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Answer» Solution :The VELOCITY at the mean position is maximum, given by `v_("max")= Aomega=1ms^(-1)` ………….(1) The ACCELERATION at the exreme position is maximum given by, `a_("max")= Aomega^(2)=1.57 ms^(-2)` ………(2) Dividing (2) by (1) we get `(a_("max"))/(v_("max"))= (Aomega^(2))/(Aomega)= (1.57)/(1)` `:. ` TIME PERIOD `T= (2pi)/(1.57)= (2(3.14))/(1.57)4s` From Eq. (1) `Aomega=1` i.e., `A(1.57)= 1` or `A= (1)/(1.57)` `:.` Amplitude A= 0.637 m |
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| 39. |
How many atoms do the molecules of a gas consist of if its gamma increases 1.2 times when the vibrational degrees of freedom are frozen? Assume the atoms to hbe linearly arranged. |
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Answer» |
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| 40. |
A bucket of mass 8 kg is supported by a light rope wound around a solid wooden cylinder of mass 12 kg and radius 20 cm free to rotate about its axis. A man holding the free end of the rope, with the bucket and the cylinder at rest initially, lets go the bucket freely downwards in a well 50 m deep. Neglecting friction, obtain the speed of the bucket and the angular speed of the cylihder just before the bucket enters water. |
| Answer» SOLUTION :`m_(1)gh = (1//2) m_(1)v^(2) + (1//2) Iomega^(2), I=(1//2) m_(2)^(2) , OMEGA = v//r , m_(1)gh = 7v^(2), v = 23.66 ms^(-1), omega = v//r = 118.3 rad s^(-1)` | |
| 41. |
According to Kepler, the period of revolution of a planet (T) and its mean distance from the Sun (R) are related by the equation |
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Answer» `T^(2)R^(3)` = CONSTANT |
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| 42. |
For a particle executing SHM. The KE K is given by K=K_(0)cos^(2)wt. The maximum value of PE is |
| Answer» Answer :A | |
| 43. |
Figure shows a square with masses 'm' corners. M.I. of the system about x-axis passing throught the centre is |
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Answer» `ML^(2)` |
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| 44. |
A temperature at which both the Fahrenheit and the centigrade scales have the same value is |
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Answer» `40^(@)` |
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| 45. |
A block of mass m is placed on a surface with a vertical cross section given by y = (x^3)/(6).If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is: |
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Answer» `1/2m` |
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| 46. |
1000 drops of a liquid each of diameter 4 mm coalesce to form a single large drop. If surface tension of liquid is 35 "dyn "cm^(-1), calculate the energy evolved by the system in the process. |
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Answer» Solution :Number of drops N = 1000 `rArrn^(1//3)=10rArrn^(2//3)=100` Surface TENSION of liquid T = 35 dyne `cm^(-1)` Radius of each small drop r = 2 mm = 0.2 cm Energy evolved in merging `W=4pir^(2)T[n-n^(2//3)]rArrW=4xx22/7xx(2XX10^(-1))^(2)35[1000-100]` `rArrW=(88xx35xx4xx10^(-2))/7[900]` = 15840 ergs. `~~1.58xx10^(-3)J` |
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| 47. |
Define angle of contact. |
| Answer» Solution :The angle between tangents drawn at the POINT of contact to the LIQUID SURFACE and solid surface inside the liquid is CALLED the angle of contact for a pair of solid and liquid. | |
| 48. |
A body thrown vertically up with velocity u reaches the maximum height h after T seconds. Which of the following statements is true ? |
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Answer» At a HEIGHT `h/2`from the ground its VELOCITY is |
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| 49. |
A block is placed on a rough horizontal plane. A time dependent horizontal force F = Kt acts on the block. Here K is a positive constant. Acceleration time graph of the block is : |
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Answer»
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| 50. |
The spring of a rocket in flight burns up due to friction .At whose expense is the heat energy required for burningobtained ? The rocket or the atmosphere ? |
| Answer» Solution :Heat energy REQUIRED for burning of casting of ROCKET comes FROMTHE rosketitself . Asresult of work done againstfrictionthe kinetic energyof rocketcontinuously decreases and this work against FRICTION REAPPEARS as heat energy . | |
