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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. |
Mercury is used as thermometric liquid of all liquids. Why? |
| Answer» Solution :Mercury is opaque and bright . It does not wet the GLASS tube. It is a good conductor of heat having UNIFORM coefficient of expansion over a WIDE RANGE of temperature . | |
| 2. |
The magnitude of displacement may or may not be equal to the path length traversed by an object." Explain this statement with example. |
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Answer» Solution :For example, for motion of the car from 0 to P, the path length is + 360 m and the displacement is + 360 m. In this case, the MAGNITUDE of displacement is EQUAL to the path length. CONSIDER the motion of the car from 0 to P and back to Q. In this case, the path length = (+ 360 m) + (+ 120 m) = + 480 m. The displacement = (+240 m) - (0 m) = + 240 m. THUS, the magnitude of displacement is not equal to the path length. 
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| 3. |
A standing wave is represented by y=A sin (100t) cos (0.01x) where y and A are in millimetres. t in seconds and x in metres. The velocity of the wave is ……… . |
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Answer» `10^(4)` m/s |
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| 4. |
The maximum tension in the string of a pendulum is two times of the minimum tension. Let theta_0 be the angular amplitude. The cos theta_0 is |
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Answer» `1/2` |
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| 5. |
A glass plate of length 10cm, breadth 1.54cm and thickness 0.20cm weighs 8.2g in air. It is held vertically with the long side horizontal and the lower half under water. The apparent weight of the plate, is found to be (W+0.179)gm.wt. Find W. Surface tension of water = 7.3 xx 10^(-2) N/m andg = 9.8m//s^2 |
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Answer» |
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| 6. |
In a horizontal capillary tube, the rate of capillary flow depends on the surface tension force as well as the viscous force. Lueas and washburn showed that the length (x) of liquid penetration in a horizontal capillary depends on a factor (k) apart from time (t). The factor is given byk = [(rTcos theta)/(2ne)]^(1//2), where r, T, theta and ne are radiusof the capillary tube, surface tension, contact angle and coefficient of viscosity respectively. If the length of liquid in the capillary grows from zero to x_(0) in time t_(0), how much time will be needed for the length to increases from x_(0) to 4x_(0). |
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Answer» |
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| 7. |
Fora simple harmonic oscillator the frequency of oscillation is independent of |
| Answer» Answer :D | |
| 8. |
You have learnt that a travelling wave in one dimension is represented by a function y =f (x,t)where x and t must appear in the combination x - vt or x + or v + vt, i.e. y = f ( x pm vt). Is the converse ture ? Examine if the following functions for y can possibly represent a travelling wave: (a) (x-vt)^(2) (b) log [ ((x + v _(t)))/(x _(0))] (c ) (1)/((x + vt)) |
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Answer» Solution :We function for one DIMENSIONAL traveling WAVE should be: `y = a sin (OMEGA t pm kx)` `=a sin {k ((omega )/(k) t pm x )}` `y =a sin (k (vt pm x)} (because v = (omega)/(k)=` wavevelocity) `implies y =f (vt pm x) OR y =f (x pm vt)` Above equation for traveling wave satisfied following condition: (i) It is continuous for all values of x and t. (ii) Its range is finite. (Here, VLAUES of y are finite in the range `-a le y le a)` (iii) It gives definite value foe given vaue of x and t. Now in present example, the functions given in OPTION (a), (b) , (c) do not satisfy above conditions because of following reasons. (a) Function `(x-vt)^(2)` has infinite range from 0 to `+ oo` (b) Function log `((x + vt)/(x _(0)))` does not give definite value for given value of x and t because it has series expansion. (c) Function `((1)/( x + vt))` is not continuous because here we can not take `x =- vt.` Thus, all the three given functions do not represent traveling wave. Hence, it is not necessary that every function `f (x pm vt) OR f (vt pm x)` should represenet travelling wave. |
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| 9. |
The radius of the earth is about 6400 km and that of mars is about 3200 km. The mass of the earth is about 10 times that of mars . If an object weights 200 N on the surfaceof earth its weight on the surface of mars would be |
| Answer» Answer :B | |
| 10. |
When 57.986 is rouned off to 4 significant figures, then it becomes |
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Answer» 58 |
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| 11. |
For a constant hydraullic stress on an object the fractional change in the object's volume (Delta V)/(V)and bulk modulus (B) are related as ........ |
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Answer» `(DELTA V)/(V) prop B` |
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| 12. |
The mass per unit length of a non - uniform rod of length L varies as m = m_(0)(x^(2))/(L), where m_(0) is a the rod is at (PL)/(4) where P is |
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Answer» |
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| 13. |
When an object is said to be a particle ? |
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Answer» SOLUTION :CONCEPT of particle is most important in mechanics. Particle means a point-like object having mass. Particle is dimensionless, so that it is ideal concept, it is not found in real. It can only perform linear motion. In following cases object can be considered as particle : (1) An object can be said to be a particle if it is moving in a straight line such that all of its particles cover same distance in same time. (2) When the dimensions of two objects arenegligible with respect to distance between them, then those objects can be considered as particles. Eg. : EARTH and SUN are considered as particles while calculating GRAVITATIONAL force between them. |
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| 14. |
Starting with the same initial conditions, an ideal gas expands from volume V_1 to V_2 om three different ways. If the work done by the gas is W_1in the isothermal process, W_2 in the isobaric process and W_3 in the adiabatic process, then |
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Answer» `W_1gtW_2gtW_3` |
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| 15. |
Which of the following is associated with liquid only and not gases? |
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Answer» Pressure |
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| 16. |
Two rods with lengths 12.321cm and 10.3 cm and placed side by side. The difference in their lengths is |
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Answer» 2.02 cm |
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| 17. |
Keeping the velocity of projection constant, the angle of projection is increased from 0^(@) to 90^(@). Then the horizontal range of the projectile |
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Answer» goes on INCREASING up to `90^(@)` |
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| 18. |
In the pulley system shown in fifure the movable pullet=ys A,B and C are of mass 1 kg eachD and E are fixed pulleys are light and inexensible. Choose the correct alternative (s). |
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Answer» TENSION in the STRING is 6.5 N |
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| 19. |
A cylinderical rod of mass .m. length L and radius .R. has two cords wound around it whose ends are attached to a rigid support. The rod is held horizontally with the two cords vertical. When the rod is released, the cords unwind and rod rotates. Find the tension in the cords as they unwound and determine the linear accelerations of the cylinder as it falls. |
Answer» Solution :The rod rotates and TRANSLATES while falling down. `MG-2T=Ma` (translational motion) `2TR = I alpha` (Rotation motion) SINCE `I=(MR^(2))/(2)` and `alpha = (a)/(R )` on solving the aboveequation we get `u=(2g)/(3)` and `T=(Mg)/(6)` |
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| 20. |
If vecA=(2hati+hatj)"and"vecB=(hati-hatj+5hatk).Find vecAxxvecB, angle between vecA"and"vecB Unit vector perpendicular to vecA"and"vecB. |
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Answer» Solution :`vecAxxvecB=|(veci,vecj,veck),(2.1,0,),(1,-1,5)|=veci(5+0)-vecj(10-0)+veck(-2-1)` `:.vecAxxvecB=5veci-10vecj-3veck, sin theta=|vecAxxvecB|/(AB)` `A = SQRT(2^(2)+1^(2))=sqrt5, B=sqrt(1^(2)+(-1)+5^(2))=sqrt(27)` `|vecAxxvecB|=sqrt(5^(2)+(-10)^(2)+(-3)^(2))=sqrt(25+100+9)=sqrt(134)` `:. sin theta = sqrt(134)/(sqrt5sqrt(27)) :. sin ^(-1)sqrt(134)/sqrt(135)` Let `vecu` be the unit vector perpendicular to `vecA "and" VECB` then `vec u` is unit vector along `vecAxxvecB`. `:. vecu= (vecAxxvecB)/|vecAxxvecB|=(5veci-10vecj-3veck)/sqrt(134)` |
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| 21. |
A wooden block of mass 20 gram and cross section 20 cm^(2) is floating on the surface of water. If it is slightly depressed and released from the equilibrium position, then calculate the frequency of oscillation of the block if specific gravity of wood is 0.28. |
| Answer» SOLUTION :0.05 HZ | |
| 22. |
ABC is a right angled trianglar plate of uniform thichness. I_(1), I_(2) and I_(3) are M.I. about AB, BC and AC respectively, Then which of the following relations is correct ? [AB = 3m BC = 4m] |
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Answer» `I_(1) = I_(2) = I_(3)` |
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| 23. |
Apply first law for (a) an isothermal (b) adiabatic (c) isobaric processes. |
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Answer» Solution :(a) For an isothermal process since TEMPERATURE is CONSTANT, the internal energy is also constant. This implies that dU or `deltaU = 0`. For an isothermal process, the first law of thermodynamics can be written as follows, `Q = W` (b) This is a process in which no heat flows into or out of the system `(Q=0)`. But the gas can expand by SPENDING its internal energy or gas can be compressed through some external work. So the pressure, volume and temperature of the system may change in an adiabatic process. For an adiabatic process, the first law becomes `DeltaU` = W. (c) The first law of thermodynamics for isobaric process is GIVEN by `DeltaU=Q-PDeltaV` `W=PDeltaV, DeltaU=Q-muRT_(f)[1-(T_(i))/(T_(f))]` |
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| 24. |
In the figure shown, upper block is given a velocity 6m//s and very long plank, velocity 3m//s The following quantities are to be matched when both attain same velocity. {:(,"Column-1",," Column-2"),("(A)","Work done by friction on 1 kg block in joule",,"(P) Positive"),("(B)","Work done by friction on 2 kg plank in joule",,"(Q) Negative"),("(C)","Magnitude of change in momentum in N-s of 2 kg plank",,"(R) 3"),("(D)","Change in KE of system consisting of block and plank in joule",,"(S) 7"),(,,,"(T) 2"):} |
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Answer» <P> |
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| 25. |
State about the correctness or otherwise of the following two statements. i) A physical quanity can have dimensions but no units. ii) A physical quantity can have units but no dimensions. |
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Answer» SOLUTION :Not CORRECT correct. |
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| 26. |
A gas with C_(P)//C_(V)=gammagoes from an initialstate (P_(1), V_(1), T_(1))to a final state(P_(2), V_(2), T_(2))through an adiabatic process. The work done by the gas is a) (nR(T_(1)-T_(2)))/(gamma-1) b) (P_(1)V_(1)-P_(2)v_(2))/(gamma-1) c) nR gamma(T_(1)-T_(2)) |
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Answer» a and B are true |
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| 27. |
Obtain the formula for the coefficient of performance of carnot engine. |
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Answer» Solution :Efficiency in CYCLIC PROCESS `eta=1-Q_2/Q_1` and efficiency of carnot engine `eta=1-T_2/T_1` `THEREFORE Q_2/Q_1=T_2/T_1 RARR Q_1/Q_2 =T_1/T_2` EXPANDING this `therefore (Q_1-Q_2)/Q_2=(T_1-T_2)/T_2` Reversing ` therefore Q_2/(Q_1-Q_2)=T_2/(T_1-T_2)` But `Q_2/(Q_1-Q_2)` is called coefficient of performance `alpha`. `therefore alpha=T_2/(T_1-T_2)` |
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| 28. |
If a_(1),a_(2),a_(3)...... a_(n) are the measured value of a physical quantity "a" and a_(m) is the true value then absolute error........ |
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Answer» `a_(m) = Deltaa_(N)+a_(n)` |
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| 29. |
A particle is projected with a velocity of 10sqrt(2)m//s at an angle of 45^(@) with the horizontal. Find the interval between the moments when speed issqrt(125)m//s. |
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Answer» SOLUTION :`(g = 10m//s^(2))` `u_(n) = 10, u_(y) = 0` `V^(2) = v_(X)^(2) + v_(y)^(2)` `125 = 100 + v_(y)^(2)` `v_(y) = 5` `Delta t = (2v_(y))/(g) = (2 xx 5)/(10) = 1s` 
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| 30. |
The rate of change of total momentum of a many -particle system is proportional to the external force/sum of the internal forces on the system . |
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Answer» <P> Solution :Internal FORCES can not change the total or NET momentum of a system . Hence, the rate of change of total momentum of many PARTICLE system is proportional to the external force on the system `(Delta p = 0p = "CONSTANT")` |
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| 31. |
The position vector of a particlevecR as a fuction of time is given by : vecR = 4sin(2 pi t)hat(i) + 4cos(2 pi i) hat(j). Where R is in meters,t is in seconds and hat(i) and hat(j) denote unit vectors along x- and y - directions, respectively. Which one of the following statements is wrong for the motion of particle? |
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Answer» Path of the particle is a circle of radius 4 METER |
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| 32. |
We can not need to apply the force an all particles of door for opening or closing it. Why? |
| Answer» Solution :Because, the relative POSITIVE REMAINS constant for all the particles of door is CONSIDERED as a rigid body and force applies to a single PARTICLE, produces the torque acting to whole door. | |
| 33. |
A woman of mass m stands at the edge of ,a solid cylindrical platform of mass M and radius R. At t = 0 the platform is rotating with negligible friction at angular velocity omega_(0) about a vertical axis passing through the centre. The woman begins to walk with speed v, relative to the platform, towards the centre of the platform: |
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Answer» Angular VELOCITY when WOMAN reaches the centre is `(V+ m/M)omega_(0)` |
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| 34. |
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional springconstant of the wire. (Torsional spring constant alpha is defined by the relation J=-alpha theta where J is the restoring couple and theta the angle of twist) |
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Answer» SOLUTION :Here m=10kg, R=15cm=0.15m,T=1.5s Moment of INERTIA of disc `I=1/2mR^(2)=1/2xx10xx(0.15)^(1)kgm^(2)` Now `T=2pisqrt(1/(alpha))` So `alpha=(4pi^(2)I)/(T^(2))=4XX(22/7)^(2)xx1/2(10xx(0.15)^(2))/((1.5)^(2))` `=1.97` Nm/rad |
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| 35. |
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant alpha is defined by the relation J=-alpha theta, where J is the restoring couple and theta the angle of twist). |
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Answer» Solution :MOMENT of inertia `I=1/2mR^(2)=0.1125kgm^(2)` TIME PERIOD T=1.5sec. Torsional spring constant `C=(4pi^(2)I)/(T^(2))(T=2pisqrt(I/C)), C=1.9` Nm/rod |
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| 36. |
For small lead balls of volumes 1cm^(3), 2cm^(3), 3cm^(3) and 4cm^(3) are descending down through the same liquid a. viscous force is more on 4(cm)^(3) ball b. terminal velocity is more for 4(cm)^(3) ball c. viscous force ismore in 1(cm)^(3) ball d. terminal velocity ismore in 1(cm)^(3) ball |
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Answer» a,B are correct |
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| 37. |
Mass of a rocket is 100 kg. When 0.1 kg fuel is burnt every second, velocity of exhaust gas coming out of it is 1Km/s.Acceleration of rocket will be ....... |
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Answer» `1000 m//s^(2)` `= (V_(G ) XX (dm)/(dt))/( m )` `=(1000 xx 0.1)/( 100)` `a= 1 m//s^(2)` |
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| 38. |
One mole of a gas at 27^(@)C and 3 atmospheric pressure is compressed to 1/3 of its volume (a) slowly (b) suddenly. What is the resulting temperature ? gamma =1 .4. |
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Answer» Solution :(a) Slowly, i.e., isothermal, PV =a constatn at constatn temperature So final temperature `= 27 ^(@)C= 300 K` (b) Suddenly i.e., adiabatic `T_(1)V_(1) ^( gamma -1) = T_(2)V_(2) ^( gamma -1) , T_(1) = 27 + 273 = 300 KV _(1) = V _(1)` `V_(2) = 1//3 V_(1) T_(2) = ?` `T_(2) = T_(1) ((V_(1))/( V _(2))) ^( gamma -1) = 300 ((V _(1))/(V _(1) //3)) ^( 1. 4 - gamma)= 300 XX (3) ^( 0.4)` `log T_(2) = log 300 + 0.4 xx log 3` `= 2. 6679 T_(2) = ` Antilog of `2. 6679` `T_(2) = 465.5K` |
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| 39. |
Moon is seen to be of (1/2)^(@) diameter from the earth. What must be the relative size compared to the earth? |
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Answer» Solution :GIVEN that moon is SEEN as `((1)/(2))^(@)` DIAMETER and earth is seen as `2^(@)` diameter. Hence`=("Diameter of earth")/("Diameter of moon")=(1)/(2^(@))=4` |
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| 40. |
The air pressure in a car tyre increases during driving. Why? |
| Answer» Solution :Due to the friction between the TYRE and the ROAD, the TEMPERATURE of tyre and hence that of air inside it INCREASES. Since the VOLUME remains constant, pressure of the air increases with the increase of temperature. | |
| 41. |
A ray of light passing through a prism having mu=sqrt2 suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. What is the angle of prism. |
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Answer» Solution :As the prism is in the position of minimum DEVIATION `delta_m=(2i-A)` with `r=A//2` According to given problem, `i=2r=A` [as `r=A//2`] `delta_m= 2A-A=A` and hence from `mu=(sin[(A+delta_m)//2])/(sin (A//2))i.e., SQRT2=(sin A)/(sin A//2) (or)` `sqrt2 sin""A/2=2 sin ""A/2 COS ""A/2 i.e., cos ""A/2=1/sqrt2 (or)` `A/2= cos^-1[1/sqrt2]=45^@ i.e., A=90^@` |
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| 42. |
In given figure, one mole of an ideal gas (gamma = 7//5) is taken through the cyclic process ABCDA. Take R = (25)/(3) J//mol -K (a) Find the temperature of the gas in states A,B,C and D. (b) Find the amount of heat supplied/released in processes AB,BC,CD and DA. (c) Find work done by gas during cyclic process. |
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Answer» (b) `3500J, 5000J, 7000J, 2500J`(c ) `-1000J` |
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| 43. |
How much should the pressure on a litre of water be changed to compress it by 0.10 % ? (B=2.2 xx 10 ^(9) N//m^(2)) |
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Answer» Solution :`B =- (p )/((Delta V )/( V ))` `THEREFORE Delta p =- B XX (Delta V )/(V) =- 2.2 xx 10 ^(9) xx (- (0.10)/(100))` `= + 2.2 xx 10 ^(6) Nm^(-2)` |
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| 44. |
To reduce friction between the moving parts of a machine, the liquid used must have |
| Answer» Answer :A | |
| 45. |
Velocity and acceleration vectors of charged particle moving perpendicular to the direction of magnetic field at a given instant of time are vartheta= 2hat(i) + c hat(j) and bar(a) = 3 hat(i) + 4 hat(j) respectively. Then the value of .c. is |
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Answer» 3 |
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| 46. |
Find the angle between two vectors with the help of scalar product . |
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Answer» SOLUTION :If the `theta ` is the angle between `vec(A) and vec(B) ` , then vector product , `vec(A). vec(B) = AB cos theta ` ` :. cos theta = (vec(A).vec(B))/(|vec(A)||vec(B)|)` ` :.cos theta = (vec(A).vec(B))/(AB)` ` :. theta = cos^(-1)((vec(A).vec(B))/(AB))` In CARTESIAN co-ordinate system , `cos theta = ((A_(X)B_(x)+A_(y)B_(y)+A_(Z)B_(z)))/((sqrtA_(x)^(2)+A_(y)^(2)+A_(z)^(2))(sqrt(B_(x)^(2)+B_(y)^(2)+B_(z)^(2))))` |
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| 47. |
Two bodies A and B of masses m_(1) and m_(2) moving with velocities V_(1) and V_(2) respectively collide. If m_(2) gt _(1) then the force exerted by B on A during the collision is |
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Answer» GREATER than the FORCE exerted by A on B |
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| 48. |
A prism of refracting angle 3^@ is made of a material of refractive index 1.652. Find its angle of minimum deviation. |
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Answer» `1.956^@` |
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| 49. |
The diameter of a sphere is measured as 1.71 cm using an instrument witha least count 0.01 cm What is the percentage error in its surface area? |
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Answer» |
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| 50. |
What is indicated by sudden fall in barometric height? |
| Answer» SOLUTION :It INDICATES the POSSIBILITY of STORM. | |