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13901.

The specific heat of a substance is 0.09 "cal/gm"^@C . If the temperature is measured on Fahrenheit scale the value of its specific heat is "cal/gm/"^@F is

Answer»

0.09
0.9
0.05
0.5

Answer :C
Discussion
13902.

In the equation (P+ (a)/(V^(2)))(V-b)= constant, the unit of a is

Answer»

<P>` "DYNE" xx cm^(5)`
` "dyne" xx cm^(4)`
` "dyne" xx cm^(3)`
` "dyne" xx cm^(2)`

Solution :As `[(a)/(V^(2))]= [P] :. [a]= [PV^(2)]`
`a= ("dyne")/(cm^(2))(cm^(3))^(2)= "dyne" xx cm^(4)`
Discussion
13903.

State whether critical velocity and terminal velocity are the same ?

Answer»

SOLUTION :Critical velocity and terminal velocity are not the same. Critical of a LIQUID is the limiting velocity for STREAMLINE flow of the liquid, but terminal velocity is the constant velocity acquired by a body MOVING through a fluid.
Discussion
13904.

A geo-stationary satellite orbits around the earth in a circular orbit which is at a height of 36000km from the surface of earth. Then the period of spy satellite orbiting a few hundred km above the earth.s surface (R_"(earth")= 6400km) will approximately be

Answer»

`1/2` HR
`1(1/2)`hr
2 hr
4 hr

Answer :B
Discussion
13905.

In the above problem angle made by velocity vector with x-axis after 1/2 second is

Answer»

`TAN^(-1)(2)`
`tan^(-1)(4)`
`tan^(-1)(1//4)`
`tan^(-1)(1//2)`

ANSWER :D
Discussion
13906.

The reduced mass of the system of two particles of masses 2 m and 4 m will be

Answer»

2
`(2)/(3) m`
`(3)/(2) m`
`(4)/(3) m`

ANSWER :D
Discussion
13907.

Convert the unit of work done from MKS system to CGS system (i.e.,joule )

Answer»

Solution :Dimensional formula of ENERGY is `M^(1)L^(2)T^(-2)`
Let IJ=n ergs rArr n =J/ergs
`:. N_(2)=n_(1)[M_(1)/M_(2)]^(1) [L_(1)/L_(2)]^(-2) [T_(1)/T_(2)]^(-2)`
`rArrn=[(1kg)/(1g)] [(1m)/(1cm)]^(2)[(1s)/(1s)]^(-2) rArrn=[(1000g)/(1g]][(100cm)/(1cm)]^(2)[(1s)/(1s)]^(-2)`
`rArrn=[1000]xx[100]^(2)xx[1]^(2) rArr n= 10^(3)xx10^(4)=10^(7)`
`:.n=10^(7) :. 1 "JOULE" =10^(7)erg`
Discussion
13908.

A cyclist pedals the cycle for some time and then stops peddling, then

Answer»

frictional force on the BACK wheel is in the direction of MOTION and FRONT wheel is OPPOSITE to the direction of motion
frictional force on the back wheel is opposite to the direction of motion and front wheel is in the direction of motion
frictional force on the both the wheels are opposite to the direction of motion.
frictional force on the both the wheels are in the direction of motion.

Answer :C
Discussion
13909.

A body of mass 0.3 kg is taken up an inclined plane to length 10 m and height 5 m and then allowed to slide down to the bottom again. The coefficient of friction between the body and the plane is 0.15. What is the (i) work done by the gravitational force over the round trip? (ii) work done by the applied force over the upward joumey? (iii) work done by frictional force over the round trip? (iv) kinetic energy of the body at the end of the trip? How is the answer to (iv) related to the first three answers?

Answer»

Solution :`sin theta = (CB)/(CA) = 0.5`
`therefore =30^@`
(i) W= FS =- mg `sin theta xx H = -14.7 J `is the W.D. by gravitational FORCE in moving the body up the inclined plane.
W. = FS + mg `sin theta xx h = 14.7`. Is the W.D. by gravitational force in moving the body down the inclined plane.
`therefore` Total W.D. round the trip, `W_1 = W + Q. = 0 `
(ii) Force needed to move the body up the inclined plane,
`F = mg sin theta + f_k = mg sin theta + mu_k R = mg sin theta + mu_k mg cos theta`
`therefore` W.D. by force over the upward journey is
`W_2 = F xx l = mg (sin theta + mu_k cos theta) l = 18.5 J`
(iii) W.D. by frictional force over the round trip,
`W_3 = - fk(l+l) = - 2 fkl = - 2 mu_k mg cos theta l = - 7.6 J `
(iv) K.E. of the body at the end of round trip.
= W.D. by net force in moving the body down the inclined plane
=(mg `sin theta - mu_k mg cos theta`) l
= 10.9 J
`IMPLIES ` K.E. of body = net W.D. on the body
Discussion
13910.

The centripetal force on a body of mass 'm' and velocity 'v' moving in circular orbit of radius 'r' is given by F = mv^2/r. Using the formula of centripetal force write an quation to find percentage error in centripetal force.

Answer»

SOLUTION :`(triangleF)/Fxx100` = (trianglem)/mxx100+2xx(triangleV)/Vxx100+(triangler)/rxx100`
Discussion
13911.

The centripetal force on a body of mass 'm' and velocity 'v' moving in circular orbit of radius 'r' is given by F = mv^2/r. Write the dimensional formula of force.

Answer»

SOLUTION :`[MLT^-2]`
Discussion
13912.

The area under velocity-time graph gives

Answer»

(a) POSITIVE
(B) NEGATIVE
(C) EITHER positive (or) negative
(d) zero

Answer :A
Discussion
13913.

A body is projected horizontally from the top of a tower. When the body strikes the ground its velocity is V and the direction of motion makes 30^(@) with the horizontal. The velocity of projection is

Answer»

`(sqrt(3)v)/(2)`
`(v)/(sqrt(3))`
`(v)/(2)`
`(2V)/(sqrt(3))`

ANSWER :A
Discussion
13914.

Surface tension of a liquid is independent of the area of the surface.

Answer»

Solution :Surface TENSION of LIQUID is the FORCE exerted by the molecules lying on one side of an imaginary line of unit length on the molecule lying on the other side of the line which is PERPENDICULAR to the line and parallel to the surface is DEFINEDAS the surface tension of the liquid.
`therefore` Surface tension `S=(F)/(l)`
The surface tension is independent of the area of the liquid surface.
Discussion
13915.

A ball of mass 150 g moving with an acceleration of 20m//s^(2) is hit by a force, which acts on it for 0.1 s. What is the impulsive force?

Answer»

0.1N
0.5N
0.3N
1.2N

Solution :Force = MASS `XX` acceleration
and Impulse `(J)= Fxxt= mxxa xx t`
`therefore J= 150xx10^(-3) xx 20 xx 0.1=300 xx 10^(-3)= 0.3N`
Discussion
13916.

Car will stop after a time (from the start)

Answer»

12.2 s
15.8 s
9.3 s
10.6 s

Answer :D
Discussion
13917.

A gras hopper can jump maximum distance of 1.6 m. It spends negligible time on the ground. How far can it go in 10 seconds ?

Answer»

Solution :`(u^(2))/(g) = 1.6 u^(2) = 16U = 4m//s`
`4 cos THETA = 4 xx (1)/(sqrt(2)) = 2sqrt(2)m//s`
`S = 4 cos theta .t = 2sqrt(2) xx 10 ""S = 20sqrt(2)m`
Discussion
13918.

The position vector of a particle is vecr = (acosomegat)hat(i)+(asinomegat)hat(j). The velocity of the particle is

Answer»

directed TOWARDS the ORIGIN
directedaway from the origin
parallel to he position VECTOR
PERPENDICULAR to the position vector.

Answer :D
Discussion
13919.

A car starts uniform motion from rest, than find shapes of graphs : (i) xtot, (ii) vto t and (iii) a tot

Answer»

Solution :(i) `x to t` GRAPH : HYPERBOLA
(ii) `V to t` graph : Straight LINE
(iii`) a to t`graph : Straight line
Discussion
13920.

According to Newton, the viscous force acting between liquid layers of area A and velocity gradient (Deltav)/(Delta z) is given by F = -eta A (dv)/(dz), where eta is constant called coefficient of viscosity . The dimensional formula of its

Answer»

`[ML^(-2) T^(-2) ]`
`[ M^(0) L^(0) T^(0) ]`
`[ML^(2) T^(-2) ]`
`[ML^(-1) T^(-1) ]`

ANSWER :A
Discussion
13921.

A planet of mass m is revolving round the sun (of mass M_(s) in an elliptical orbit. If vecv is the velocity of the planet when its position vector from the sun is vecr, then areal velocity of the positon vector of the planet is :

Answer»

`VECV + VECR`
`vecrxx vecv`
`1/2(vecv xxvecr)`
`1/2(vecrxx vecv)`

ANSWER :D
Discussion
13922.

A charge particle of charge 'q' and mass 'm' enters in a given magnetic field 'B' and perpendicular to the magnetic field as shown in the figure. It enters at an angle of 60^(@) with the boundary surface of magnetic field and comes out at an angle of 30^(@)with the boundary surface of the magnetic field as shown in the figure. Find width 'd' in which magnetic field exist.

Answer»

`((sqrt3-1)MV)/(4qB)`
`((sqrt3-1)mv)/(QB)`
`((sqrt3-1)mv)/(3qB)`
`((sqrt3-1)mv)/(2qB)`

ANSWER :d
Discussion
13923.

In Fig. a massive rod AB is held in horizontal position by two massless strings. If the string at B breaks and if the horizontal acceleration of centre of mass, vertical acceleration and angular acceleration of rod about the centre of mass are a_x,a_y respectively, then

Answer»

`2sqrt(3)a_(y)=SQRT(3)alphal+2a_(x)`
`sqrt(3)_(y)=sqrt(3) alphal+a_(x)`
`a_(y)=sqrt(3)alphal+2a_(x)`
`2a_(y)=alphal+2sqrt(3)a_(x)`

SOLUTION :Acceleration of POINT `A` along the string is ZERO. THEREFORE
`a_(x)cos30^@-a_(y)cos60^@+(alphal)/2cos60^@=0`
Discussion
13924.

What is the pressure on a swimmer 10 m below the surface of a lake ? (Density of water = 10^(3)kgm^(-3) , Acceleration of gravity g=10ms^(-2) and atm pressure P_(a)=1.01xx10^(5)Pa)

Answer»

Solution :Depth from the surface of lake H= 10 m
Density of water `rho=10^(3)kgm^(-3)`
Acceleration of gravity `g=10ms^(-2)`
Atm pressure `P_(a)=1.01xx10^(5)Pa`
Pressure at a depth h from the surface of lake water,
`P=P_(a)+hrhog`
`=1.01xx10^(5)+10xx10^(3)xx10`
`=1.01xx10^(5)+10^(5)`
`=2.01xx10^(5)NM^(-2)or 1.99 atm`.
Hence , this is a `100%` increase in pressure fromsurface level . At a depth of 1 km the increase in pressure is 100 atm . Submarines are designed to WITHSTAND such enormouspressure .
Discussion
13925.

If theacceleration due to gravity on earth is 9.81 m//s^2 and the radius of the earth is 6370 km fing the mass of the earth ? (G=6.67xx10^(-11) Nm^2//kg^2)

Answer»

Solution :`g=9.81 m//s^2`, R=6370 km = `6.370xx10^6` m
`G=6.67xx10^(-11) Nm^2//Kg^2`
`g=(GM)/R^2 RARR` Mass of the EARTH `M=(gR^2)/G`
`=(9.81(6.37xx10^6)^2)/(6.67xx10^(-11))= 5.97xx10^24` kg .
Discussion
13926.

Three particles of masses 1 kg, 2 kg and 3 kg are situated at the corners of an equilateral triangle of side b in the x-y plane with mass 1 kg at the origin and 2 kg on the x-axis. The coordinates of the centre of mass are

Answer»

`((7B)/(12) , (3 SQRT3 b)/(12))`
`((3 sqrt3 b)/(12) , (7b)/(12))`
`((b)/(12) , (3 sqrt3 b)/(12))`
`((7b)/(12) , (sqrt3b)/(12))`

SOLUTION :The coordinates of points A , B and C are (0 , 0) (b , 0) and `((b)/(2) , (bsqrt3)/(2))` respectively
`X_(CM) = (1 xx 0 + 2 xx b+ 3 (b //2))/(1 + 2 + 3) = (7b)/(12)`
`Y_(CM) = (1 xx 0 + 2 xx 0 + 3sqrt3 (b //2))/(1 + 2 + 3) = (3sqrt3 b)/(12)`
So , the coordinates of centre of MASS are `((7b)/(12) , (3 sqrt3b)/(12))`
Discussion
13927.

Red shift is an illustration of …….. .

Answer»

LOW temperature emission
high frequency ABSORPTION
DOPPLER effect
Same UNKNOWN Phenomenon.

Solution :Doppler effect
Discussion
13928.

The diameter of a soap bubble is 8 mm and the surface tension of soap solution is 4xx10^(-3)Nm^(-1). Find the excess pressure inside the bubble.

Answer»

SOLUTION :Diameter of th BUBBLE `8mm=8xx10^(-3)m`
Radius of the bubble `(r)=4xx10^(-3)m`
Surface tension (S) `=4xx10^(-2)NM^(-1)`
Excess pressure inside the bubble `=(4S)/r`
`=(4xx4xx10^(-2))/(4xx10^(-3))=40Nm^(-2)`
`:.` Excess pressure inside the bubble `=40Nm^(-2)`
Discussion
13929.

{:(,"ColumnI",, "ColumnII"),((A),"Gas A is ... and gas B is ...",(p),"Monoatomic, diatomic"),((B),p_(A)//p_(B) "is",(q),"Diatomic, monoatomic"),((C),n_(A)//n_(B)"is",(r), gt 1),(,,(s), lt1),(,,(t),"cannot say anything"):}

Answer»

<P>

ANSWER :`(ATOR,BTO"p",Ctoq,DTOS)`
Discussion
13930.

From length of square plate and force acting on it pressure on plate is measured. If error in force and length is 4% and 2% respectively, then maximum relative error in measurement of pressure will be …………..

Answer»

<P>0.01
0.02
0.06
0.08

Solution :`P=(F)/(A)`
For square PLATE `A=L^(2)`
`P=(F)/(L^(2))`
`:.` PERCENTAGE error in measurement of pressure,
`=deltaFxx100%+2deltaLxx100%`
`=4%+2xx2%`
`=4%+4%=8%`
Discussion
13931.

A glass vessel contains air at 60^(@)C. To what temperature must it be heated to expel one third of the air, the pressure . remaining constant. (Neglect the expansion of the vessel)

Answer»

`127^(@)C`
`226.5^(@)C`
`327^(@)C`
`427^(@)C`

ANSWER :B
Discussion
13932.

If vecA=3hati+4hatj"and"vecB=7hati+24hatj, find a vector having the same magnitude as vecBand parallel to vecA

Answer»

SOLUTION :The vector parallel to `VECA `and having magnitude of `vecB" is "vecC=|vecB|vecA/|vecA|=|vecB|hatA`
`B=SQRT(7^(2)+24^(2))=25"and"hatA=hatA/A=(3hati+4hatj)/sqrt(3^(2)+4^(2))=1/5(3hati+4hatj)`
`vecC=25xx1/5(3hati+4hatj)=15hati+20hatj`
Discussion
13933.

A circular platform is mounted on a vertical frictionless axle. Its radius is r = 2m and its moment of inertia is I = 200 kg -m^(2). It is initially at rest. A 70 kg man stands on the edge of the platform and begins to walk along the edge at sped V_(0)=1.0 m//s relative to the ground. Find the angular velocity of the platform.

Answer»

SOLUTION :Angular MOMENTUM of MAN = angular momentum of platform in opposite direction.
or `mv_(0)=I omega rArr omega = 0.7` rad/s.
Discussion
13934.

Which has more Young.s modulus, a thin steel wire or thick steel wire

Answer»

thin
thick
both have same
none of the above

Answer :3
Discussion
13935.

1 revolution is equivalent to 360^(@). The value of 1 revolution per minute is

Answer»

`2PI ra//s`
`0.104rad//s`
`3.14 rad//s`
None of these

Solution :(b) `1/("rev")//("min")=(1xx2pirad)/(60s)=(PI)/30 rad//s=0.1047rad//s`
Discussion
13936.

Explain why (a) The angle of contact of mercury with glass is obtuse , while of water with glass is acute .

Answer»

Solution :Let a drop of a LIQUID (L) is on a solid surface S PLACED in AIR A. The air is the medium around drop . The surface tensions are SHOWN on drop .

`S_(SL)=` The surface tension corresponding to solid -liquid layer .
`S_(LA)` = The surface tension corresponding to liquid - air layer.
`S_(SA)` = The surface tension corresponding to solid - air layer.
`theta` = angle of contact
`S_(LA)costheta+S_(SA)=S_(SL)`
For the mercury glass interface , `S_(SA)ltS_(SL)`Therefore , `costheta` is negative . Thus `theta` is an acute angle.
Discussion
13937.

An ice box is built of wood of thickness 1.75 cm. The box has an inner lining of cork 2 cm thick. If the difference in temperature between the interior of the box and outside is 12°C, calculate the temperature of the interface between wood and cork. Thermal conductivity of wood and cork are respectively 0.25 and 0.05 in S.I. unit.

Answer»


Solution :`0.25 XX A xx (285 - theta) // 1.75 xx 10^(-2)= 0.05 xx A ( theta - 273)//2 xx 10^(-3) , theta = 283.21` K
Discussion
13938.

Which of the following is a vector quantity?

Answer»

ANGULAR momentum
Work
Potential energy
Electric current

Answer :A
Discussion
13939.

Show that, if three forces acting on a particle can be taken sequentially to form the three sides of a triangle, theirresultant is zero.

Answer»

Solution :Let `veca , vecb and vecc` be the three forces acting at a point O [Fig.2.73 (a)].Now, TAKING these forces sequentially and KEEPING their magnitudesand directionsthe same, we obtain the three sides of a triangle, `vec(PQ) , vec(QR) and vec(RP)` [Fig.2.73(b)].

We have to show that `veca+vecb+vecc=vec0`
According to the triangle LAW of vector addition,
`veca+vecb=vec(PR)=-vec(RP) or, veca+vecb=-vecc`
or, `veca+vecb+vecc=vec0`
`therefore` The resultant of the forces `veca,vecb and vecc` is ZERO.
Discussion
13940.

A particle moving with uniform retardation covers distances 18 m. 14 m and 10 m in successive seconds. It comes to rest after travelling a further distance of

Answer»

50m
8m
12m
42m

Answer :B
Discussion
13941.

Absolute temperature can be calculated by

Answer»

MEAN square VELOCITY
MOTION of the molecule
Both (a) and (b)
None of the above

ANSWER :A
Discussion
13942.

A particle experience constant acceleration for 20 seconds after starting from rest. If it travels a distance s_(1) in the first 10 seconds and distance s_(2) in the next 10 seconds, then

Answer»

`s_(2) = s_(1)`
`2s_(2) = s_(1)`
`s_(2) = 3s_(1)`
`s_(2) = 4s_(1)`

Answer :C
Discussion
13943.

Figures 14.25 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure. Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.

Answer»

Solution :(a) x = - 3 sin `pit` where x is in cm.
(B) x = - 2 cos `pi/2t` where x is in cm.
Discussion
13944.

The earth moving around the sun in a circular orbit is acted upon by a force and hence work must be done on the earth. Do you agree with this statement?

Answer»

Solution :The statement is WRONG. The earth REVOLVES around the sun under the force of attraction of the sun. This force (centripetal) is always perpendicular to the MOTION of the earth. Therefore, `theta=90^(@)` and `W=FS cos 90^(@)=0`. HENCE, sun does to work on the earth.
Discussion
13945.

A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t =0 to 12 s.

Answer»

Solution :Here initial speed `v_(0)=0 h = 90 m`
Lost speed in each collision, `=1/10 xx ` Speed acquired while striking ground,
If the velocity of ball while striking ground is `v_(1),` then
`V _(1) ^(2) - v _(0) ^(2) = 2gh`
`v _(1) ^(2-) -0 = 2 xx 9.8 xx 90`
`v _(1) ^(2) sqrt (2 xx 9.8 xx 90)`
` v _(1) = 42 ms ^(-1)`
and time taken to reach the ground is `t_(1), `then
From `d = v_(0) t + 1/2 G t ^(2) `
`h = 1/2 g t _(1) ^(2)`
`t _(1) = sqrt ((2g)/(g))`
` = sqrt ((2 xx 90)/(9.8))`
`t _(1) = 4.28 s`
If the velocity when returning back after striking the ground is `v _(2),` then
`v _(2) = v _(1) - (10 % of v _(1))`
`= 42 -4.2`
`= 37.8 ms ^(-1)`
Height achieved by ball after striking ground is `h_(1),` then
`h _(1) = (v _(2) ^(2))/( 2g) = (37.8 xx 37.8)/(2 xx 9.8) = 72.9m`
Now when ball again strikes the ground, velocity is `V_(3),` then its displacemtn will be O.
So, `v_(3) ^(2) - v _(2) ^(2) = 2g xx 0 [ because h =0]`
`therefore v _(3) ^(2) = v _(2) ^(2)`
`v _(3) = v _(2) = 37.8 ms ^(-1)`
If time taken between two successive colisions is `t _(2),` then
`0= v _(2) t _(2) + 1/2 (-g) t_(2) ^(2)`
`0= 37.8 t _(2) - (9.8)/(2) t _(2) ^(2)`
`4.9 t _(2) ^(2) = 37.8 t _(2)`
`t _(2) = 7.714 s.`
If velocity after second collision is `v _(4),` then
`v _(4) = v _(3) - 10% of v _(3)`
`= 37.8 -3.78`
`v _(4) = 34.02 ms ^(-1)`
Now `t _(2) = 7.714s` is time of flight (2T)
`[because` Time taken to reach maximum height]
={Time taken to return back}
So that `t _(2) = 2t.`
`t. = (t_(2))/(2)`
` t . = ( 7.714)/(2) =3.857=3.86s`
Now `t_(1) + t _(2) =4.28 + 7.714=11.994 s=12s`

Thus, ball will return back in `t =0` to t =12 s after colliding ground.
Velocity of ball is `37.8 ms ^(-1)` when strikes first time on ground.
Discussion
13946.

The motion represented by the eqution y(t)=Acos(Omega+phi) is called SHM.The displacement y(in cm) of an oscillating particle varies with time t (in second) according to the equation y=2cos(0.5pit+pi/3).Find the amplitude and period of partice.

Answer»

SOLUTION :y=Acos`(omega+phi)` GIVENY= 2COS `(0.5pit+pi/3)`
Discussion
13947.

A vessel contains oil (density = 0.8g//cm^(2) over mercury (density = 13.6g//cm^(2). A uniform sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of sphere in g//cm^(2) is

Answer»

`3.3`
`6.4`
`7.2`
`12.8`

ANSWER :C
Discussion
13948.

A glass slab of thickness 4cm contains the same number of waves as 5cm of water when both are traversed by the same monochromatic light . IF the refractive index of water is 4//3 then that of glass is

Answer»

`5//3`
`5//4`
`16//15`
`3//2`

ANSWER :A
Discussion
13949.

In the following cyclic process is The above process in the P-T coordinates is given as

Answer»




SOLUTION :AB and CD are isobaric process so PRESSURE will be CONSTANT
BC and DA are isochoric process so volume will be constant
Also, by PV=nRT
P must be proportional to T if volume is constant and PASSING through origin
Discussion
13950.

The height of which a cylindrical vessel of radius R be filled with a homogeneous liquid to make the force with which the liquid presses on the side of the vessel equal to the force exerted by the liquid on the bottom of vessel is

Answer»

`H=R`
`h=(R)/(2)`
`h=(R)/(3)`
`(R)/(9)`

ANSWER :A
Discussion