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13301.

Express 1 HP in terms of watts.

Answer»

SOLUTION :`1H.P=746.3W~~746W`
Discussion
13302.

A small sphere freely falling from a certain height after striking the floor rebounds and rises to same height. When is it possible ?

Answer»

SOLUTION : A.Here `E = sqrt (( H _(2))/( h _(1))) =1,`This is POSSIBLE when the collision is CLASTIC.
Discussion
13303.

The inability of objects to move on its own or change its state of motion is called as

Answer»

FORCE
MOMENTUM
INERTIA
impulse

SOLUTION :inertia
Discussion
13304.

Four spheres A, B, C and D of different metals but of same radius are kept at same temperature. The ratio of their densities and specific heats are 2 : 3 : 5 : 1 and 3 : 6 : 2 : 4. Which sphere will show the fastest rate of cooling (initially) :

Answer»

A
D
C
B

ANSWER :D
Discussion
13305.

Effective length of a simple pendulum is 50 cm and mass of the bob suspended is 4 g. The bob is pulled to one side to make the string horizontal and then it is released. What will be the kinetic energy of the bob when it makes an angle 60^(@) with the vertical?

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ANSWER :98000 ERG
Discussion
13306.

The pitch of a screw gauge is 1mm and there are 100 divisions on its circular scale. When nothing is put in between the studs and the two studs are in contact as well, the zero of the circular line lies 6 divisions above the reference line. When a thin uniform wire is placed firmly between the two studs, 2 divisions on the pitch scale / linear scale are clearly visible, while 62^(nd) division on the circular scale coincides with the reference line. The diameter of the wire is

Answer»

2.56 MM
2.68 mm
2.62 mm
2.02 mm

ANSWER :B
Discussion
13307.

Ball A is dropped from the top ofa building at the same instant that a ball B is thrown vertically upward from the ground . When the balls collide, they are moving in opposite directions and the speed of A is twice the speed of B . At what fraction of the height of the building did the collsion occur?

Answer»

Solution :Given `V_(A)=2V_(B)`
Let `h_(1)"and"h_(2)` are the DISTANCES travelled by the two balls
`:. sqrt(2gh_(1))=2sqrt(u^(2)-2gh_(2))`
`2gh_(1)=4U^(2)-8gh_(32), 2gh_(1)+8gh_(2)=4u^(2), 2g[h_(1)+4h_(2)]=4u^(2)`
`h_(1)+4h_(2)=(2u^(2))/g` ...(1)
Again as `V_(A)=22V_(B),`from `v=u+at o+"gt" =2(u-"gt")`
`:."gt"=2u-2"gt" rArr 2u=3"gt"`
`:. t=2u//3G` ...(2)
`h_(1)+h_(2)=ut=u(2u/3g)=(2u^(2))/(3g) `...(3)
`h_(1)+4h_(2)=(2u^(2))/g` ...(4)
solving (3)&(4) we GET `h_(1)=(2u^(2))/(9g),h_(2)=(4u^(2))/(9g)`
`:. h_(1)/h_(2)=(2u^(2)//9g)/(4u^(2)//9g)=1/2`
Discussion
13308.

What will be the change in phase of wave dueto reflection from free support?

Answer»

Solution :When a tankeraccelerates a PSEUDO force EXERTED on oil and hence its surface does not REMAIN horizontal .
Consider a small ELEMENT of mass dm . The forces on it are shown as in figure .
For equailibrium , the force parallel to the slope, `(dm)acostheta=(dm)gsintheta)`
where the pseudo force exerted opposite to the acceleration (dm)a
`thereforeacostheta=gsintheta`
`thereforetantheta=(a)/(g)`=slope
Discussion
13309.

A man of 50 kg is standing at one end on a boat of length 25 m and mass 200 kg. If the starts running and when he reaches the other end, he has a velocity 2ms^(-1) respect to the boat. The final velocity of the boat is (in ms^(-1))

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`2//5`
`2//3`
`8//5`
`8//3`

ANSWER :A
Discussion
13310.

Rain is falling vertically with a speed of 20ms^(-1). A person is running in the rain with a velocity of 5ms^(-1) and a wind is also blowing with a speed of 15ms^(-1) (both from the west). The angle with the vertical at which the person should hold his umbrella so that he may not get drenched is:

Answer»

Solution :`vec(V_("Rain"))=vec(V_(R))=20(-hat(K))`
`vec(V_("MAN"))=vec(V_(M))=5hat(i)`
`vec(V_(wind))=vec(V_(W))=15hat(i)`
Resultant velocity of rain and wind =
`vec(V_(RM))=-20hat(K)+15hat(i)`
Now, Velocity of Rain relative to man=
`vec(V_(RM))-vec(V_(M))`
`=(-20hat(K)+15hat(i))-(5hat(i))`
`=-20hat(K)+10hat(i)`
`Tanalpha==(1)/(2)rArr alpha = Tan^(-1)(1)/(2)`
Discussion
13311.

A calorimeter whose water equivalent is 5xx10^(-3) kg is filled with 50xx10^(-3) kg of water at 80^(@)C. The temperature falls to 75^(@)C in 4 minutes. When it is filled with 40xx10^(-3)kg of another liquid, the time to fall through same range of temperature (from 80^(@)C to 75^(@)C ) is 130 seconds. Find the specific heat capacity of the liquid.

Answer»


ANSWER :`2603Jkg^(-1)K^(-1)`
Discussion
13312.

0.2 m^(3) of an ideal gas at a pressure of 2 Mpa and 600 K is expanded isothermally to 5 times the initial volume. It is then cooled to 300 K at constant volume and then compressed back polystro-pically to its initial state. Determine the work done and heat transfer during the cycle.

Answer»

Solution :Considering process 1-2 which is isothermal, i.e., `T_(1)=T_(2)=600K`
` P_(2)= (p_(1)V_(1))/V_(2)=(2xx10^(6)xx0.2)/(5xx0.2)=400xx10^(3) N//m^(2)=0.4 MPA`

Work done, `W_(i-2)=p_(1)V_(1) log_(e). V_(2)/V_(1)=2xx10^(6)xx0.2 log_(e) 5 = 643775 J= 643.775 kJ`
Considering process2-3 which is at constant volume,
i.e., `V_(2)=V_(3)=5xx0.2=1 m^(3)`
`P_(3)=P_(2)xxT_(3)/T_(2)=400xx10_(3)xx300/600 = 200xx10^(3) N//m^(2)=0.2 MPa`
`W_(2-3)` = 0 as volume remains constant.
For the polytropic process `3-1 =P_(3)V_(3)^(n)=p_(1)V_(1)^(n)`
`n=(log_(e).(p_(1))/(p_(3)))/(log_(e). V_(3)/V_(1))=(log_(e). 2/0.2)/(log_(e). 1/0.2)=2.3026/1.6094=1.433`
`W_(3-1)=(p_(3)V_(3)-p_(1)V_(1))/(n-1)=(0.2xx10^(6)xx1-2xx10^(6)xx0.2)/(1.433-1)`
`=-461894 J=-461.894 kJ`
Net work done, `W_(n)=W_(1-2)+W_(2-3)+W_(3-1)=643.774+0+(-461.894)=181.88 kJ`
:. Heat transfer during the complete cycle=181.88 kJ
Discussion
13313.

We are familiar with Newton's lows of motion. A circular racetrack of redius 300 m is banked at on angle of 15^@ . The coffecient of friction between the wheels of a race car and the road is 0.2. Find: The optimum speed of the race car to avoid wear and tear on its tyrrs.

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Solution :REFER Solved Problem EXAMPLE 22.
Discussion
13314.

A ball is thrown stright upwards with a speed v from a point h meters above the ground .Find the time taken for the ball to strike the ground.

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`V/g [1+sqrt(1+(2HG)/(v^(2)))]`
`v/g [1-sqrt(1-(2hg)/(v^(2)))]`
`v/g [1-sqrt(1+(2hg)/(v^(2)))]`
`v/g [2+(2hg)/v^(2)]`

ANSWER :A
Discussion
13315.

Four rods of same material but with different radii and lengths are used to connect two reservoirs of heat with the same temperature difference. Which one will conduct more heat

Answer»

r = 1 cm, 1 = 1M
r = 1CM, 1 = 2m
r = 1cm, `1=(1)/(2)` m
`r=(1)/(2)` cm, `1=(1)/(2)` m

Answer :C
Discussion
13316.

Three particles of masses 8 kg, 4 kg, and 4 kg situated at (4, 1), (-2, 2) and (1, -3) are acted upon by external forces 6bar(i)N, -6 bar(i)N and 14 bar(i)N. The aceleration of centre of mass of the system is

Answer»

`0.625 MS^(-2)`
`6.25 ms^(-2)`
`2.2 ms^(-2)`
`22 ms^(-2)`

ANSWER :A
Discussion
13317.

When direction of object moving on circular path is changed what will be change in direction of motion

Answer»

SOLUTION :DIRECTIONOF centripetalforcedo notdependon directionof motionhencethere WILLNOT beanychangeby changingdirection
Discussion
13318.

Discuss the forces acting on the n particles in a system.

Answer»

Solution :The force acting on centre of mass,
`MvecA=vec(F_(1))+vec(F_(2))+...vec(F_(n))" here "vec(F_(1))` is not only one force but it is vector sum of forces acting only a PARTICLE.
Similarly for second, third, … .n. particle the vector sum of force actin are `vec(F_(2))+vec(F_(3))+...vec(F_(n))`
The forces acting on particles of a system are of two types :
(i) Internal forces due to mutual interactions among the particles and
(ii) External forces
In the FIGURE (a) a system made of particle-1 and particle-2 is shown.

External forces acting on particles 1 and 2 are `vec(F_(1))andvec(F_(2))` respectively and mutual forces interaction are denoted by `vec(F_(12)) and vec(F_(21))`.
Here, all the forces may be considered as acting on the centre of mass which is shown in figure (b).
According to Newton.s third law of motion the internal forces being equal and opposite their resultant is zero. So the force `vecF` acting on the centre of mass can be REGARDED as resultant external forces only.
`therefore MvecA=vec(F)_(ext)` where `vec(F)_(ext)-vecF` is the resultant force of only external forces.
This equation indicates that the system moves under the effect of resultant external force as if the whole mass of the system where concentrated as its centre of mass.
Discussion
13319.

Consider an expression F = A x sin^(-1)'(Bt) where F represents force, x represents distance and t represents time. Dimensionally the quantity AB represents

Answer»

energy
surface TENSION
intensity of light
pressure

Solution :Dimensional analysis SUGGESTS that the quantity Ax on RHS MUST have the DIMENSIONS of force whereas B must have the dimensions of recipocal of time. Then, the product AB will have the dimensions of energy per unit area per unit time, the same as those of intensity of light.
Discussion
13320.

A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute SHM with a time period. T= 2pi sqrt((m)/(A p g)) where, m is mass of the body and p is density of the liquid. .

Answer»

Solution :Suppose, LOG is pressed DOWNWARD to y, the volume of liquid displaced by block will be yA.
The mass of displaced water `M= Vp`
`=yA p""[therefore y xx A= " volume V "]`
Buoyant force the upward by displaced water = weight of displaced water
`therefore F= -Mg""`(Weight and Buoyant force are opposite to each other)
`= -y A pg`
`therefore = -(A pg )y`
`therefore F propto -y`
where `Ap g =k` constant
HENCE, force acting on log is directly proportional to the displacement and opposite it. So motion of log is SHM
Now, period of SHM particle `T= 2PI sqrt((m)/(k))`
but `k= A p g`
`therefore T= 2pi sqrt((m)/(A p g))` is PROVED.
Discussion
13321.

The average energy of a molecules of a monoatomic gas at temperature T is (K=Boltzmann constant )

Answer»

`1/2kT`
KT
`3/2kT`
`5/2kT`

ANSWER :C
Discussion
13322.

Represent instantaneous displacement of a wave motion.

Answer»

Solution :y ( X,t ) = `a SIN ( omega t - KX + phi)` where ` omega //k = 'V'` is the velocity of the wave and k- is propogation constant of the wave.
Discussion
13323.

A source and a listener are approaching closer with a relative velocity of 40 ms^(-1).If the true source frequency is 1200 Hz, calculate the observed frequency under these conditions : (i)The source along is moving (ii)The listener along is moving Take speed of sound in air 340 ms^(-1).

Answer»


ANSWER :1360 HZ, 1341 Hz
Discussion
13324.

During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.

Answer»

Solution :(a) False : Frictional force acts OPPOSITE to the direction of motion of centre of mass of a BODY. In the case of rolling, the direction of motion of the centre of mass is backward, hence, frictional force acts in the forward direction.
(b) True : Rolling can be considered as the ROTATION of a body about an axis passing through the POINT of contact of the body with the ground. Hence, its instantaneous speed is zero.
(C ) False : Since body is rolling, it has contripetat force and its instantaneous acceleration is non-zero. So it is false.
(d) True : When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, work done against friction is also zero.
(e) True : The rolling of a body occurs when a frictional force acts between the body and the surface. The frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body steps from the slope under the effect of own weight.
Discussion
13325.

A rocket works on the principle of conservation of

Answer»

energy
MASS
ANGULAR MOMENTUM
LINEAR momentum

Solution :mass
Discussion
13326.

linear expansion is change in length of an object with temperature. The absolute zero is _____. (-273.15^@C,-273.15K,-273.15^@F, 0^@C

Answer»

SOLUTION :`-273.15^@C`.
Discussion
13327.

The displacement time graph of moving particle is shown below. The instantaneous velocity of the particle is negative at the point.

Answer»

D
F
C
E

Answer :D
Discussion
13328.

If a body of mass 36gm moves with S.H.M of amplitude A = 13 cm and period T = 12sec . At a time t = 0 the displacement is x = +13cm. The shortest time of passage from x = + 6.5 cm to X = -6.5 is

Answer»

4sec 
2 SEC 
6 sec
3 sec 

Answer :B
Discussion
13329.

What is smelting ? Explain the process with example.

Answer»

Solution :The inability of objects to move on its own or change its STATE of motion is called inertia. Inertia means resistance to change its state. There are three types of inertia:
1. Inertia of REST: The inability of an object to change its state of rest is called inertia of rest. E.g. (i) When a stationary bus starts to move, the passengers experience a sudden backward push. (ii) A book lying on the table will remain at rest until it is moved by some external agencies.
2. Inertia of motion: The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion. E.g. (i) When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front SEAT. (ii) An athlete running in a race will continue to run even after reaching the finishing point.
3. Inertia of direction: The inability of an object to change its direction of motion on its own is called inertia of direction. E.g. (i) When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle. (ii) When a bus moving along a straight line TAKES a turn to the right. The passengers are THROWN towards left.
Discussion
13330.

A block is pushed up a rough inclined plane of 45^(@) with a velocity u. The block rises up and then comes down sliding to have a velocity u/2 at the bottom. The coefficient of sliding friction between block and the plane is

Answer»

`0.5`
`0.6`
`0.8`
`0.45`

ANSWER :B
Discussion
13331.

A ball with an initial momentum 'P' collides normally with a rigid wall. If P_(1) is the linear momentum after the perfectly elastic collision ,………….

Answer»

<P>`P_(1) = P`
`P_(1) = -P`
`P_(1) = -2P`
`P_(1) = 2P`

SOLUTION :`P_1 = -2P`
Discussion
13332.

An L-shaped tube is held in a stream. It has a small orifice at the upper end which is 0.106 m above the surface of water. If water in stream is flowing at 2.45 ms^(-1) and entering at the other end of the tube, then find the height of the jet of water coming out from the orifice.

Answer»

Solution :We know that `v_(1)^(2)-v_(2)^(2)=2/p(P_(2)-P_(1))=2/p(hrhog)=2hg`
`=2 xx 0.106 xx 9.8 =2.0776=2.08`
HENCE, `v_(2)^(2)=2.45^(2)-2.08^(2)`
`therefore v_(2)=1.98ms^(-1)`
We GET `0=(1.98)^(2)-2(1.98)H`
Height to which jet REACHES will be, `h=(1.98 xx 1.98)/(19.6)=0.20m`
h=20.0m
Discussion
13333.

A satellite of mass m revolves around the earth R at a height x from its surface. If g is the acceleratioon due to gravity on the surface of the earth, the orbital speed of the satellite is

Answer»

`SQRT(GX)`
`sqrt((GR)/(R-x))`
`sqrt((gR^(2))/(R-x))`
`sqrt((gR^(2))/(R+x))`

ANSWER :D
Discussion
13334.

If an air bubble rises from the bottom of a mercury lank to the top its volume becomes 1.5 times. When normal pressure is 76cm of Hg then the depth of the Hg tank is

Answer»

38 CM
132 cm
76 cm
49 cm

Answer :A
Discussion
13335.

When the terminal velocity is reached, the acceleration of a body moving through a viscous medium is ……………. .

Answer»

ZERO
POSITIVE
NEGATIVE
EITHER (b) or (c ) depending upon other factors

Answer :A
Discussion
13336.

In whirling motion, if the string is cut suddenly, the stone moves tangential to circle is an example for

Answer»

INERTIA of MOTION
Inertia of DIRECTION
Inertia of rest
back pull

Solution : Inertia of direction
Discussion
13337.

Coefficient of friction between two blocks shown inFig. 7.203 is mu = 0.4 . The blocks are given velocities of 2 m/s and 8 m/s in the directions shown in figure . The timewhen relative motion between them will stop is ( in sec) . :

Answer»


Solution :Using principle of conservation of MOMENTUM:
`1 XX (2)+ 2 xx (8) = 3V implies v = 6 m//s `
For 1 kg block , `f = mu m_(1) g= 0.4 XX1 xx 10 = 4 N`
`THEREFORE "" a = (f)/(m_(1)) = (4)/(1) = 4 ms^(-2)`
Using , `"" v_("rel") = u_("rel") + at `
6= 2 + (4) t
t = 1 second
Discussion
13338.

If force (F), velocity (v) and time (T)are taken as fundamental units, the dimension of mass are .....

Answer»

`[FV^(-1)]`
`[FVT^(-2)]`
`[FV^(-1)T^(-1)]`
`[FV^(-1)T]`

Solution :`m=kF^(a)V^(b)T^(c) .....(1)`
Where K is dimensionless CONSTANT and a, b,CER comparing dimension of both sides,
`M^(1)L^(0)T^(0)=[M^(1)L^(1)T^(-2)]^(a)xx[M^(0)L^(1)T^(1)]^(c)`
`=M^(a)L^(a)T^(-2a)xxL^(b)T^(-b)xxT^(c)`
`M^(1)L^(0)T^(0)=M^(a)L^(a+b)T^(-2b-b+c)`
Comparing POWERS of M,L,T
`a=1,a+b=0 "" -2a+c=0`
`:.1+b=0, "" -2xx1-(-1)+c=0`
`:.b=-1 "" :. -2+1+c=0`
`:. c=1`
`:. m=F^(1)v^(-1)T^(1)` [From equ. (1)]
Discussion
13339.

As shown in fig. a block of mass m = 0.40 kg is moving on a frictionless surface at a speed of v = 0.50 m//s.A block of force constant k = 750 N /m comprised the spring and becomes rest for a instant . How much spring will be compressed ?

Answer»

SOLUTION :`1.2 xx10^(-2) m or 1.2 CM`
Discussion
13340.

A cylinder rests on a horizontal rotating disc, as shown in the figure. Find at what angular velocity, omega, the cylinder falls off the disc, if the distance between the axes of the disc and cylinder is R, and the coefficient of friction mugtD//h where D is the diameter of the cylinder and It is its height.

Answer»


SOLUTION :The centripetal force that keeps the cylinder at rest on the disc is the frictioal force `f`. According to a non inertial observer on disc a pseudo force the cylinder reacts with an equal and oppostite force, `F`, which sometimes is reffered to as the centrifugal force.

`F=MOMEGA^(2)R`
where `M` is the mass of the cylinder.
The cylinder can fall off either by slipping away or by tilting about point `P`, depending of whichever takes place first. the critical agular speed `w_(1)` for slipping occurs when `F` equals `f:F=f`
`Momega_(1)^(2)R+mugM`
where `g` is the gravitational acceleration. HENCE `omega_(1)=sqrt((mug)/R)`
`F` TIES to rotaste the cylinder about `P`, but the weight `W` OPPOSES it. The rotatiion becomes pssible, when the torque caused by `W`.
`Fh/2=W D/2implies F=W D/h`
`Momega_(2)^(2)R=Mg D/h`
giving `omega_(2)=sqrt(D/(hR))`
Since we are given `mugtD/h`, we see that `omega_(1)gtomega_(2)` and the cylinder falls off by rolling over at `omega=omega_(2).`
Discussion
13341.

In the figure shown , a light ring with three rods, each of mass m is welded on this. The rods from an equilateral triangle. The rigid assembly is released on rough fixed inclined plane. The minimum value of the coefficient of static friction, that will allow pure rolling of the assembly 1/x tantheta where 'x' is

Answer»


ANSWER :2
Discussion
13342.

A refrigerator is working between the temperature of melting ice and room temperature, which is 290 K. Calculate work done by the refrigerator to freeze 2 kg of water.

Answer»

SOLUTION :Work done ` = Q _(1) - Q _(2) = (T_(1) - T _(2)) Q _(2) // T _(2) = (290 - 273) xx 2 xx 333 xx 10 ^(3) // 273 = 4.14 xx 10 ^(4) J`
Discussion
13343.

Calculate the height at which the value of acceleration due to gravity becomes 50% of that at the surface of the earth. (Radius of the earth =6400 km )

Answer»

Solution :g at HEIGHT , `g_(H)=g/((1+h/R)^2)`
In this PROBLEM , `g_(h)=50/100g = 0.5 g`
`0.5g =g/((1+h/R)^2)IMPLIES(1+h/R)^2=g/(0.5g)=2`
`1+h/R=sqrt2,h/R=sqrt2-1implies` the height ,
`h = 0.414 R = 0.414 xx 6400 = 2650` km.
Discussion
13344.

A spring balance reads 10 kg when a bucket of water is suspended from it. What is the reading on the spring balance when an ice cube of mass 1.5 kg is put into the bucket

Answer»

`22.5kg`
`2.25 KG`
`1.15 kg`
`11.5kg`

ANSWER :D
Discussion
13345.

A body projected vertically from the earth reaches a height equal to earth's radius returning to the earth. The power exerted by the gravitational force is greatest

Answer»

at the instant just before the BODY hits the EARTH
it REMAINS constant all the through
at the instant just after the body is projected
at the HIGHEST position of the body

Answer :A
Discussion
13346.

Derive the expression of Kinetic energy in rotation.

Answer»

Solution :(i) Let us consider a rigid body rotating with angular VELOCITY `omega` about an axis as shown in Figure.
(ii) Every particle of the body will have the same angular velocity `omega` and different tangential velocities v based on its positions from the axis of rotation.

(iii) Let us choose a particle of mass `m_(i)` situated at distance `r_(i)` from the axis of rotation. If has a tangential velocity `v_(i)` given by the relation, `v_(i)=r_(i)omega`. The kinetic energy `KE_(i)` of the particle is,
`KE_(t)=1/2m_(i)v_(i)^(2)`
(iv) Writing the expression with the angular velocity,
`KE_(t)=1/2m_(i)(r_(i)omega)^(2)=1/2 (m_(i)r_(i)^(2))omega^(2)`
(v) For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with summation as,
`KE=1/2(summ_(i)r_(i)^(2))`
(vi) where, the term `summ_(i)r_(i)^(2)` is the moment of inertia I of the whole body. `I=summ_(i)r_(i)^(2)`
(vii) Hence, the expression for KE of the rigid body in ROTATIONAL motion is,
`KE=1/2Iomega^(2)`
This is analogous to the expression for kinetic energy in TRANSLATIONAL motion.
`KE=1/2Mv^(2)`
(VIII) Relational between rotational kinetic energy and angular momentum
Let a rigid body of moment of inertia I rotate with angular velocity `omega`.
The angular momentum of rigid body is,
`L=Iomega`
(ix) The rotational kinetic energy of the rigid body is, `KE=1/2Iomega^(2)`
(x) By multiplying the numerator and DENOMINATOR of the above equation with I, we get a relation between Land KE as,
`KE=1/2(I^(2)omega^(2))/I=1/2((Iomega^(2)))/I`
`KE=L^(2)/(2I)`
Discussion
13347.

If f = x^2 then the reative error in f is

Answer»

`[(2/_\X)/x]`
`(/_\x)/x`
`(/_\x)/x`
`/_\x^2`

ANSWER :A
Discussion
13348.

The velocity of a body is given by the equation v= b//t + ct^(2) + dt^(3). Find the dimensional formula for b.

Answer»

Solution :`(b//t)` should have the dimensions of VELOCITY
b has the dimensions of (velocity `xx` TIME)
`[b] = [LT^(-1)][T] = [L] = [M^(0)L^(1)T^(0)]`
Discussion
13349.

Suppose an object is thrown with initial speed of 10 ms^(-1) at an angle pi//4 with the horizontal, what is the range-covered? Suppose the same object is thrown similarly in the moon, will there be any change in the range? If yes, what is the change? (The acceleration due to gravity in the moon g_("moon")=1//6 g)

Answer»

SOLUTION :In projectile motion, the RANGE of particle is given by,
`R=(U^(2)SIN2THETA)/(g)`
`theta=pi//4u=v_(o)=10ms^(-1)`
`R_("earth")=((10)^(2)sinne//2)/(9.8)=100//9.8`
`R_("earth")=10.20m` (Approximately 10 m)
If the same object is thrown in the Moon, the range will increase because in the Moon, the acceleration due to gravity is smaller than g on Earth,
`g_("moon")=(g)/(6)`
`R_("moon")=(u^(2)sin2theta)/(g_("moon"))=(v_(0)^(2)sin2theta)/(g//6)`
`:.R_("moon")=6R_("earth")`
`R_("moon")=6xx10.20=61.20m`(Approximately 60 m)
The range attained on the Moon is approximately SIX times that on Earth.
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13350.

A cork and a metal bob are connected by a string as shown in the figure. If the beaker is given an acceleration towards left, then the cork will be thrown towards.

Answer»

RIGHT
left
upwards
downwards

ANSWER :B
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