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12951.

Different types of equilibrium are given in column I. Match the types given in column I with conditions given in column II.

Answer»

i-iii, 2- I, 3-v, 4-vi
1-vi,2-v,3-I,4-iii
1-I,2-ii,3-iii,4-iv
1-v,2-iii,3-I,4-ii

Answer :B
Discussion
12952.

A Steel rod of length 25 cm has a cross - sectional area of 0.8cm^(2). The force required to stretch this rod by the same amount as the expansion produced by heating it through 10^(@)C is (coefficient of linear expansion of steel is 10^(-5)//^(@)C and Young's modulus of steel is 2xx10^(10)N//m^(2)).................... .

Answer»

(a) 40 N
(b) 80 N
(c) 120 N
(d) 160 N

Solution :The REQUIRED force is GIVEN by
`F=YA ALPHA Deltat=2xx10^(10)xx0.8xx10^(-4)xx10^(-5)xx10`
`F=160N`
Discussion
12953.

A body of weight w_(1) is suspended from the ceiling of a room through a chain of weight w_(2). The ceiling pulls the chain by a force

Answer»

`w_(1)`
`w_(2)`
`w_(1)+w_(2)`
`(w_(1)+w_(2))/(2)`

ANSWER :C
Discussion
12954.

A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The crosssectional area of the wire is 0.065 cm. Calculate the elongation of the wire when the mass is at the lowest point of its path. [Y_("Steel") =2 xx 10 ^(11) N,m ^(-2)]

Answer»

Solution :Centripetal force in circular motion ,
`F = (mv ^(2))/(r ) ""omega = 2pi f=2pi xx 2 `
`= mr omega ^(2) [ because v = r omega ]""=4pi rads^(-1)`
When it is at te lowest position of the vertical circle total force,
F= weight + centripetal force
`= mg + mr omegga ^(2) = m [g + r omega ^(2)]`
`=14.5 [9.8 + 1 xx (4 xx 3.14 ) ^(2) ] [because r =l =1 m]`
`=14.5 [9.8 +(12.56) ^(2) ]=14.5 xx 167.55 = 2492.5`
Young.s modulus `Y = ("Stress")/("Strain") = ((F )/(A))/((Delta l )/(l ))`
`therefore Deltal = (Fl )/(AY) = (2429.5 xx 1 )/(65xx10^(-7) xx 2 xx 10 ^(11))`
`therefore Delta l = 18.688 xx 10 ^(-4) m`
`therefore Delta l = 1.87 xx 10 ^(-3) 10 ^(-3) m = 1.87 mm`

Second Method:
`l = 1m`
`m = 14.5 KG`
`omega = 2 ("Rotation")/(s) = 2 xx 2 pi (rad)/(s) = 4pi (rad)/(s)`
`A = 0.065 cm ^(2) = 0.065 xx 10 ^(-4) m ^(2)`

When the body is at the lowest position of the circle, then the centripetal force,
`(mv ^(2))/( l ) = T - mg `
`therefore T = mg + (mv ^(2))/( l )`
`therefore T = mg + (ml ^(2) omega ^(2))/( l)`
`therefore T = mg + mlomega ^(2) = 14.5 xx 9.8 + 14.5 xx 1 xzx (4pi ) ^(2) = 2433.9 N`
Now `Y = (sigma )/(sum)=( F //A)/( (Delta L )/(l ))`
`Y= (FL )/( A Delta l ) = (Tl )/( A Delta l )`
`therefore Delta l =(Tl )/(AY) = ( 2433.0 xx 1 )/( ( 0.065 xx 10 ^(-4)) xx 2xx 10 ^(11))`
`therefore Delta l = 1.87 xx 10 ^(-3) m = 1.87 mm`
Discussion
12955.

A closed pipe and an open pipe emit fundamental tones of the same frequency. Find out the ratio oftheirlengths.

Answer»


Answer :Letthe LENGTH of the closed pipe and the open pipe be `l_(1) and l_(2)` , respectively . V is the velocity of sound in air.
Forthe closed pipe, the fundamental frequency is ` n_(1) = (V)/(4l_(1))` ;
for the open pipe, it is `n_(2) = (V)/(2l_(2))`
Inthis problem, ` n_(1) = n_(2) `or,`(V)/(4l_(1)) = (V)/(2l_(2)) `or,`(l_(1))/(l_(2)) = (1)/(2)`
Therefore, the ratio of their LENGTHS is 1 : 2 .
Discussion
12956.

Which of the following graph shows the variation of surface tension with temperature over small temperature ranges for water?

Answer»




ANSWER :B
Discussion
12957.

What is the purpose of using shockers in a car?

Answer»

SOLUTION :To decrease the IMPACT of force by INCREASING the time for which force acts.
Discussion
12958.

Two wheels A and C connected by a belt B as shown in the figure. The radius of C is three the radius of A. What would be the ratio of the rotational inertias (I_(C))/(I_(A)) of both the wheel have the same rotational kinetic energy ?

Answer»


ANSWER :9
Discussion
12959.

Show that the torque acting on a body is equal to the product of moment of inertia and the angular acceleration of the body.

Answer»

SOLUTION :HOLLOW SPHERE, because M. I in GREATER.
Discussion
12960.

A block of weight 8 kg is resting on a smooth horizontal plane. If it is struck by a jet of water at the rate of 2 kg s^(-1) and at the speed of 20 m s^(-1) , then the initial acceleration of the block is

Answer»

`15 MS^(-2)`
`10MS^(-2)`
`2.5ms^(-2)`
`5 ms^(-2)`

SOLUTION :`5 ms^(-2)`
Discussion
12961.

In case of compression, isothermal curve lies…..the adiabatic curve. Fill in the blanks

Answer»

above
below
sometimes above and other TIME below
cannot say

Solution :`PV= "CONSTANT" , PV^(GAMMA)= "constant"`
Discussion
12962.

The gravitational potential changes uniformly from-20 J /kg to - 40 J/kg as one moves along X-axis from x =- 1 m to x = + 1 m . Mark the correct statement about gravitational field intensity of origin.

Answer»

The gravitational field intensity ATX = 0 must be equal to 10 N/kg.
The gravitational field intensity atx = 0 may be equal to 10 N/kg.
The gravitational field intensity atx = 0 may be greater THAN10 N/kg.
The gravitational field intensity atx = 0 must not be LESS than10 N/kg.

Answer :B::C::D
Discussion
12963.

A body of mass m falls from a height h_(1) and rises to a height h_(2). The magnitude of the change in momentum during the impact with the ground.

Answer»

`mg(h_(1)+h_(2))`
`m(sqrt(2gh_(1))+sqrt(gh_(2)))`
`m(sqrt(2gh_(1))-sqrt(2gh_(2)))`
Zero

ANSWER :B
Discussion
12964.

A block of mass m = 0.1 kg is released from a height of 4m on a curved smooth surface. On the horizontal surface, path AB is smooth and path BC offers coefficient of friction mu = 01. If the impact of block with the vertical wall at C be elastic, the total distance covered by the block on the horizontal surface before coming to rest will be (take g = 10 m//s^(2))

Answer»

29 m
49 m
59 m
109 m

Answer :C
Discussion
12965.

In a normal adult, the average speed of the blood through the aorta (radius r = 0.8 cm) is 0.33 ms^(-1) . From the aorta, the blood goes into major arteries, which are 30 in number, each of radius 0.4 cm. Calculate the speed of the blood through the arteries.

Answer»

Solution :`a_1v_1 = 30 a_2v_2 implies PI r_1^2 v_1 = 30 pi r_2^2 v_2`
`v_2 = 1/30((r_1)/(r_2))^(2) v_1 implies v_2 = 1/30 xx ((0.8 xx 10^(-2)m)/(0.4 xx 10^(-2) m))^(2) xx (0.33 MS^(-1))`
`v_2 = 0.044 ms^(-1)`.
Discussion
12966.

Which of the following is a null vector (a) velocity vector of a body moving in a circle with a uniform speed (b) velocity vector of a body moving in a straight line with a uniform speed (c) position vector of the origin of the rectangular coordinate system (d) displacement vector of a stationary object

Answer»

both a & b
both b & C
a, b & c
both c & d

Answer :A
Discussion
12967.

How is the swimmer jumping into water from a height be able to make loop in air ?

Answer»

Solution :He makes use of LAW of conservation of ANGULAR momentum i.e, `I omega` is a constant `I_(1)omega_(1) = I_(2) omega_(2)`.When he pulls his arms and legs INWARD, his moment of inertia decreases and angular velocity increases.
Discussion
12968.

If the point of suspension of a simple pendulum is in a horizontal motion, with constant acceleration, what will be the effect on the time period?

Answer»

Solution :When the point of suspension moves horizontally with a constant acceleration a, the effective acceleration due to gravity with RESPECT to the point of suspension,
`G.=sqrt(g^(2)+a^(2))`
`therefore` Time period, `T=2pisqrt(L/((g^(2)+a^(2))^(1//2)))""...(1)`
From EQUATION (1), it is understood that the time period DECREASES due to the increase in the value of accleration of the bob.
Discussion
12969.

A plane is inclined at an angle 30^(@) with horizontal. The component of a vector vec(A) = - 10 hat(K) perpendicular to this plane is: (here z-direction is vertically upwards)

Answer»

`5 SQRT2`
`5 SQRT3`
5
2.5

Answer :B
Discussion
12970.

At what angle with the horizontal should a base ball be thrown so as to hit the farthest point ?

Answer»

`45^@`
`60^@`
`90^@`
`120^@`

ANSWER :A
Discussion
12971.

In case of beats, produced by the superposition of two waves of equal amplitudes, if maximum intensity is x times the intensity of superposing wave then x = ……..

Answer»

Solution :AMPLITUDE of resultant wave in beats is,
`A. =2A cos {((omega _(1) - omega _(2))/(2)) t }`
`IMPLIES (A.) _(max) = 2A`
`(because cos {( (omega _(1) - omega _(2))/(2) )t }` has maximum VALUE equal to 1)
Now, `I _(max) prop (2A) ^(2)`
Also we knwo that `I prop A ^(2)`
`THEREFORE (I _(max))/(I) = (4A ^(2))/(A ^(2)) =4 ""...(1)`
But, as per the statement `I _(max)=xI,`
`therefore (I _(max))/(I) =x""...(2)`
From equations (1) and (2),` x =4`
Discussion
12972.

Two force 10 N each inclined to each other at angle 120^(@) act on a 6 kg mass at rest. Its kinetic energy after 12 s is

Answer»

400 J
1200 J
600 J
900 J

Answer :B
Discussion
12973.

A ring of mass M and R is rotating about its axis with angular velocity omega. Two identical bodies each o mass m are now gently attached at the two ends of a diameter of the ring. Because of this the kinetic energy loss will be :

Answer»

`(m(M+2m))/(M)OMEGA^(2)R^(2)`
`(MM)/((M+m))omega^(2)R^(2)`
`(Mm)/((M+2m))omega^(2)R^(2)`
`((M+m)M)/((M+2m))omega^(2)R^(2)`

Answer :C
Discussion
12974.

Which of the following examples represent periodic motion ?

Answer»

A SWIMMER completing one (return) trip from one bank of a river to the OTHERAND back.
A FREELY suspended BAR magnet displaced from its N-S DIRECTION and released.
A hydrogen molecule rotating about its centre of mass.
An arrow released from a bow.

Answer :A::B::C::D
Discussion
12975.

A rocket isprojected vertically upwards from the surface of the earth (radius =R) with a velocity v. To what height will the rocket rise ?(Neglect air friction)

Answer»

`h=R/((2gR)/(v^2)-1)`
`h=R/((2gR)/(v^2)+1)`
`h=R((2gR)/(v^2)-1)`
`h=R((2gR)/(v^2)+1)`

ANSWER :A
Discussion
12976.

The coefficient of linear expansion for a certain metal varies with temperature asalpha = aT .IfL_(0)=1 is the initial length of the metal and the temperature of metal is charged from 27^@C to 127^@C,then the change in Internal energy of the system is (where alpha = 2 xx10^(-5)K^(-2) )

Answer»

L = 1.15 m
L = 1.70 m
L = 0.3 m
L = 1.51 m

ANSWER :A
Discussion
12977.

Aparticle oscillating under a force vecF=-k vecx- b vec v is a (k and b are constants)

Answer»

linear oscillator`
damped oscillator
forced oscillator
simple harmonic oscillator

SOLUTION :A particle oscillating under a force `VECF=- k vec x- b vecv` is a damped oscillator. The firstterm `- kvecx` represents the resotring force and SECOND term `-b vecv` represents the damping force.
Discussion
12978.

A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0cm of water in one arm and 12.5cm of spirit in the other. What is the specific gravity of spirit.

Answer»

`0.8`
`1.8`
`2.2`
`0.4`

ANSWER :A
Discussion
12979.

(A) The blue colour of sky is on account of scattering of sunlight. ( R) The intensity of scattered light varies inversely as the fourth power of wavelength of light.

Answer»

Both A and R are TRUE and R is the correct EXPLANATION of A
Both A and R are true and R is not the correct explanation of A
A is true and R is false
Both A and R is false

Answer :A
Discussion
12980.

A steel ball of mass 0.1 kg falls freely from a height of 10m an bounces to a height of 5.4m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (specific heat of steel=460JKg^(-1)K^(-1)) (g=10ms^(-2)).

Answer»

`0.01^(@)C`
`0.1^(@)C`
`1^(@)C`
`1.1^(@)C`

ANSWER :B
Discussion
12981.

The internal energy of an ideal gas decreases by the same amount as the work done by the system

Answer»

The process MUST be ADIABATIC
The process must be isothermal
The process must be isobaric
The TEMPERATURE must decrease

Answer :A::D
Discussion
12982.

A particle moves along the X-axis as x=u(t-2)+a(t-2)^2 (a) the initial velocity of the particle is u (b) the acceleration of the particle is a (c) the acceleration of the particle is 2a (d) at t = 2 s particle is at the origin

Answer»

a and B are correct
b and C are correct
a and c are correct
c and d are correct

Answer :D
Discussion
12983.

A man is hanging from a rope attached to a balloon containing hot air. The system is at rest in air. If the man climbs up the rope to the balloon, then centre of mass of the system

Answer»

remains at REST
moves upwards
moves DOWNWARDS
FIRST moves up and then moves back to the initial position

ANSWER :A
Discussion
12984.

The volume of a gas expands by 0.25 m^(3) at a constant pressure of 10^(3) N/m, the workdone is equal to .........

Answer»

250 W
2.5 W
250 N
250 J

Solution :WORKDONE = P. `DELTA V = 10^(3) xx 0.25 = 250 J`
Discussion
12985.

The components of a particles velocity in the directions at right angles are 3 m//s,4 m//s "and "12m//s respectively . The actual velocity of the particle , in m//s is

Answer»


ANSWER :13
Discussion
12986.

A freely falling body takes 2 seconds to reach the ground on a planet, when it is dropped form a height of 8m . If the perod of a simple pendulum is pie seconds on the planet the length of pendulum is (cm).

Answer»

100
50
25
200

Answer :A
Discussion
12987.

A spring of force constant k is cut in to lengths of ratio 1:2:3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k". Then k':k" is:

Answer»

`1:14`
`1:6`
`1:9`
`1:11`

ANSWER :D
Discussion
12988.

Why are tyres made of rubber not of steel?

Answer»

Solution :SINCE COEFFICIENT of friction between rubber and ROAD is less than the coefficient of friction between steel and road.
Discussion
12989.

The internal energy of an ideal gas decreasesby the same amount as the work done by the system a) The process must be adiabatic b) The process must be isothermalc) The process must be isobaricd) The temperature must decrease

Answer»

only a,d are CORRECT
only B, d are correct
only C, d are correct
only b,c are correct

Answer :A
Discussion
12990.

…………… is used for measurig long time intervals of the order of ……….. Sec.

Answer»


ANSWER :RADIOACTIVE DATING ; `10^(17)`
Discussion
12991.

Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10% of energy. How much power is generated by the turbine ? (g = 10 m/s^2)

Answer»

12.3 kW 
7.0 kW 
8.1 kW 
10.2 kW 

Solution :Mass of water falling/second = 15 kg/s.
`H = 60 m, g = 10 m//s^2` , loss = 10% ie., 90% is used
Powre generated = `15 XX 10 xx 60 xx 0.9 = 8100 W = 8.1 kW`.
Discussion
12992.

Define Orthogonal unit vectors.

Answer»

SOLUTION :`hat(i) , hat(j)and hat(k)`be THREE unit VECTORS that specify the directions along positive x-axis, positive y-axis and positiive z-axis RESPECTIVELY. These are called orthogonalunit vectors .
Discussion
12993.

Which of the following quantities does not remain constant in a uniform circular motion?

Answer»

SPEED
momentum
kinetic ENERGY
mass

Answer :B
Discussion
12994.

A particle moves in a circular path such that its speed v varies with distance as v = alpha sqrt(s) where alpha is a positive constant. Find the acceleration of paarticle after trafversing a distance S?

Answer»

`ALPHA^(2)SQRT((1)/(4)-(S^(2))/(R^(2)))`
`alpha^(2)sqrt((1)/(4)+(S^(2))/(R^(2)))`
`alpha sqrt((1)/(2)+(S^(2))/(R^(2)))`
`alpha^(2) sqrt((1)/(2)+(S^(2))/(R^(2)))`

Answer :B
Discussion
12995.

When a stone falling from the top of vertical tower has fallen a distance xm,another is let fall from a point y m below the top .If the fall from rest and reach the ground together,then the height of the tower is

Answer»

`((X+y)^2)/(4X) m`
`(4(x+y)^2)/(x) m`
`(4x)/((x+y)^2 ) m`
`4x(x+y)^2 m`

ANSWER :A
Discussion
12996.

Three moles of an ideal gas (C_p = 7/2 R)at pressure P_A and temperature T_Ais isothermally expanded to twice its initial volume. It is then compressed at constant pressure to its original volume. Finally the gas is heated at constant volume to its original pressure P_A . The correct P-V and/or P-T diagrams indicating the process are

Answer»




ANSWER :A
Discussion
12997.

A body is falling freely under the action of gravity alone in vacumm . Which of the following quantities remain constant during the fall ?

Answer»

KINETIC ENERGY
Potential energy
TOTAL mechanical energy
Total LINEAR momentum

Answer :C
Discussion
12998.

A rod AB of length 1m is placed at the edge of a smooth table as shown in fig. It is hit horizontally at point B. If the displacement of centre of mass in 1 s is 5sqrt(2) m. The angular velocity of the rod is 5x rad/s where x is (g=10" m/s"^(2))

Answer»


ANSWER :6
Discussion
12999.

The intensity of uniform magnetic field between poles of a magnetic is H. If a straight wire carrying current is placed in between the poles. The direction of force on the wire is-

Answer»

PERPENDICULARTO both CURRENT and MAGNETIC FIELD
in the DIRECTION of current
same as magnetic field
in all directions

Answer :A
Discussion
13000.

The momentum of a body is doubled. By what percentage does its kinetic energy increase?

Answer»

<P>

Solution :`K= (p^(2))/(2m) rArr (K_(1))/(K_(2))= (p_(1)^(2))/(p_(2)^(2))`
Let `k_(1)= K, p_(1)= p " then " p_(2)= 2p, K_(2)`=?
`(K)/(K_(2)) = (p^(2))/((2p)^(2)), K_(2)= 4K`
% increase in kinetic energy= `("Increase in kinetic energy")/("Initial kinetic energy")XX 100`
`(K_(2)-K_(1))/(K_(1)) xx 100 = (4K-K)/(K) xx 100= 300%`
Discussion