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12101.

Two pendulums with identical bobs and lengths are suspended from a common support such that in real position the two bobs are in contact (figure). One of the bobs is released after being displaced by 10^(@) so that it collides elastically head-on with the other bob. (a) Describe the motion of two bobs. (b) Draw a graph showing variation in energy of either pendumlum with time, for 0 le t le 2T , where T is the period of each pendulum.

Answer»

Solution :(a) Consider the adjacent diagram in which the bob B is displacedthrough an angle`theta ` and released
At t = 0 , suppose bob B is DISPLACED by `theta = 10^(@)` to the right , It is GIVEN potential energy `E_(1) =E . `Energy of A , `E_(2)=0`
WhenB is released , it strikes A at t = T/4 .In the head - on elastic collision between B and A comes to rest and A gets VELOCITY of B . Therefore`e_(1) = 0 and E_(2) = E ` Att = 2T/4 ,B reachesitsextreme right position whenKE of A isconvertedinto `PE = E_(2)=E ` Energy of B , `E_(1) = 0 `
At t = 3T/4 , A reachesits MEAN position , when its PE is converted into `KE = E_(2) =E` . It collideselastically with B and transfers whole of its energy to B . Thus , `E_(2) = 0 and E_(1) = E ` . The entire process is repeated .

(b) The value of energies of B and A at different TIME intervals are tabulated here . the plot of energy with time ` 0 le t le 2T ` is shownseparately for B and A in the figure below .
.
Discussion
12102.

A wire 5 m long and cross section 10^(-4)m^(2) is fixed at one end and a mass of 1200 kg is hung from the other end. Find the elongation (y=0.98xx10^(11) N//m^(2))

Answer»

Solution :`Y=(MGL)/(pir^(2)(DELTAL))=(Mgl)/(A(Deltal))=(1200xx9.8xx5)/(10^(-4)XX(Deltal))=0.98xx10^(11)`
`:.Deltal=6xx10^(-3)m=6 MM`
`:.` elongation =6 mm
Discussion
12103.

A perfect gas is contained in a cylinder kept in vacuum. If the cylinder suddenly bursts, then the temperature of the gas

Answer»

In increased
Becomes ZERO K
Remains unchanged
In decreased

Answer :D
Discussion
12104.

Reading of temperature may be same on

Answer»

CELSIUS and KELVIN scale
FAHRENHEIT and Kelvin scale
Celsius and Fahrenheit scale
all the three scales

Answer :B::C
Discussion
12105.

There is a rectangular metal plate in which two cavities in the shape of rectangle and circle are made, as shown with dimensions. P and Q are centres of these cavities. On heating the plate, which of the following quantities increase?

Answer»

`pi R^2`
`AB`
`R`
`B`

Answer :A::B::C::D
Discussion
12106.

In nature the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by (Ypi r ^(2))/(4R) Yis the Young's modulus, r is the radius 4R of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

Answer»

Solution :Let us SEE the diagram of the given situation ACCORDING to the problem the bending torque on the trunk of radius r of tree `= (Ypi r ^(4))/( 4R)`
where R is the radius of curvature of the bent SURFACE,

when trunk is bend then, `Wd = (Y pi r ^(4))/(4R)`
`If R GT gt H,` then the centre of gravity of tree is at a height `l = 1/2 h` from the ground.
From `Delta ABCR ^(2) = (R - d)^(2) + ((1)/(2) h ) ^(2)`
`If d lt lt R, R ^(2) = R ^(2) - 2 Rd + 1/2 h ^(2)`
`therefore d = (h ^(2))/(8R)`
If `omega _(0) ("weight")/("volume") ,` then
`(Ypi r ^(4))/( 4R) = omega _(0) ( pi r ^(2) h ) ( h ^(2))/( 8R)`
`[because ` Torque is caused by the weight]
`therefore h = ((2Y)/( omega _(0))) ^(1//3)r ^(2//3)`
Hence, critical height `=h= ((2Y)/(omega _(0))) ^(1//3)r ^(2//3)`
Discussion
12107.

(vec(a) - vec(b)) xx (vec(a)+ vec(b)) is equal to

Answer»

0
`vec(a) XX vec(b)`
`2(vec(a) xx vec(b))`
`|a|^(2) + |b|^(2)`

Answer :C
Discussion
12108.

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to gravity at the surface the planet is equal to that at the surface of the earth. If the radius of the earth is R , the radius of the planet would be .......

Answer»

`1/2R`
2R
4R
`1/4R`

Solution :`implies` ACCELERATION of gravitation `g = (GM)/R^2`
where mass `m = d xx (4)/3 piR^3` , where d = density
`:.g = G4/3 dpiR`
From Formula,
For earth : `g_e = G(d) piR_e g G, 4/3 pi` is constant
For planet : `g_p = G (2d) pi R_p:.d R = `constant
Now , `g_e =g_p implies dR_e = 2dR_p`
`:. d_eR_e = d_p R_p`
`:. R_p = R_e / 2` But `R_2 = R and dp = 2de `
`:. R_p = R/2`
Discussion
12109.

(A):The displacement of a body may be zero though its distance can be finite (R ):If a body moves such that finally it arrives at initial point ,then displacement is zero while distance is finite

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 but (R) is false
Both (A) and (R) are false

Answer :A
Discussion
12110.

A pair of physical quantities having the same dimensional formula is

Answer»

FORCE and work
Work and energy
Force and torque
Work and power

Answer :B
Discussion
12111.

Distinguish between average velocity and average speed .

Answer»

Solution :Average VELOCITY: The average velocity is defined as ratio of the displacement VECTOR to the corresponding timeinterval.
`vec(V) _("avg")= (Delta vec(r))/(Delta t)`
It is a vector quantity . The direction of average velocity is in the direction of the displacement vector `(Delta vec(r))` .
Average speed :THEAVERAGE speed is defined as the ratio of total path length travelled by the particle in a time interval.
`{:("Average"),("Speed"):}}= {{:("Total path"),("length"),(/"total time"):}`
It is a scalar quantity . doesn't possess direction.
Discussion
12112.

Which one, crown glass or flint glass, has a larger refractive index ?

Answer»

Solution :FLINT glass has larger refractive INDEX.
Discussion
12113.

An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15^@.What is the radius of the loop ?

Answer»

Solution :If N is the normal force on the wings,
`N COS theta = mg, N SIN theta = (mv^2)/( R)`
which give`R = (v^2)/(G tan theta) = (200 xx 200)/(10 xx tan 15^@) = 15 KM`
Discussion
12114.

If DeltaH=mL, where m is mass of body. DeltaH= total thermal energy supplied to the body L= latent heat of fusion. Find the dimensions of latent of fusion.

Answer»

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

SOLUTION :(B) `DeltaH=mL :[L]=([ML^(2)T^(-2)])/([M])=[L^(2)T^(-2)]`
Discussion
12115.

Write equation of centripetal acceleration and centripetal force for uniform circular motion.

Answer»

Solution :Whenbody OFMASS m moveson circularpath of radiusr withconstantspeedv ithascentripetalor radialacceleration `a= (v^(2))/( R )`. Directionofaccelerationis towardcentreof thecirclewhichis SHOWNINFIGURE

bynewtons secondlawof motionnecessaryforcetoprovidethisacceleration`f_(c ) =(mv^(2))/(R )`
In differentsituationcentripetal forceisprovidedasfollows
(1)Forplanetrevolvingaroundthe sunnecessarycentripetalis providedbygravitationalforce.
(2) Forelectron revolvingarounduncleus in atomnecessary centripetalforce isprovidedbycoulombain force
(3)forvehicles movingon level CIRULAR tracknecessarycentripetalforceis provided byfrictionforcebetweentyeandroad.
Discussion
12116.

The SHM is represented by y= 3sin 314 t+ 4 cos 314t y in cm and t in second. Find the amplitude of SHO.

Answer»

Solution :`y= 3sin 314t + 4 cos 314t` shows SHM.
We can have displacement of SHM at time t from the quation `y= A SIN (OMEGA t+phi)`.
`y= A sin omega t cos phi + A cos omega t sin phi`
In `y= (A cos phi) sin omega t+(A sin phi) cos omega t" putting "omega = 314 rad s^(-1)`.
COMPARING equation `y= (A cos phi) sin 314 t+ (A sin phi) cos 314t` with given equation `y= 3sin 314 t + 4 cos 314t`.
`3= A cos phi " and "4= A sin phi"""........"(1)`
`therefore (3)^(2) +(4)^(2) = A^(2) cos^(2) +A^(2) sin^(2) phi`
`therefore 9+ 16 =A^(2) (cos^(2) phi+ sin^(2) phi)`
`therefore A^(2)= 25 implies A = pm 5 cm`.
Now from equation (1),
`(4)/(3) = tan phi`
`therefore phi = tan^(-1) (1.3333)`
`phi = 53.8^(@)`
and periodic time `T= (2pi)/(omega)`
`=(2xx 3.14)/(314) implies T= 0.02 s`
and maximum VELOCITY, `v_("max") = A omega = 5xx 314`
`therefore v_("max") = 1570 cm"/"s`.
Discussion
12117.

A prism of glass is dipped in to water as shown in the figure. If the refractive index of water is 4//3 then the incident ray will be totally reflected if

Answer»

`SIN THETA GT 8/9`
`sin theta LT 8/9`
`sin theta=9/8`
`sin theta=8/9`

ANSWER :A
Discussion
12118.

The terminal velocity of a copper ball of radius 2 mm falling through a tank of oil at 20^@C is 6.5 cm s^(-1). Computethe viscosity of the oil at 20^@ C . Density of oil =1.5 xx 10^(3) kg m^(-3) density of copper is 8.9 xx 10^(3) kg m^(-3).

Answer»

SOLUTION :`eta = 2/9 (r^2)/V(rho - rho.) g = (2 xx (2 xx 10^(-3))^(2) xx (8.9 xx 10^3 - 1.5 xx 10^3) xx 9.8)/(9 xx 6.5 xx 10^(-2))`
`eta = 0.992` DECAPOISE
Discussion
12119.

A body displaced 10m under the force of 10 N . If the work doneon the body is 50 J , then the anglebetween the force and displacement will be ……..

Answer»

`120^(@)`
`90^(@)`
`60^(@)`
None of these

SOLUTION :`W = vec(F).vec(d) = FD COS theta `
` cos theta = W/(F.d)`
` :. K . = 0.125 J `
` :. K/k = (0.125)/(2.5)`
` :. (K-K)/K xx100 = (2.5 -0.125)/(2.5) xx100`
` = (2.375 )/(2.5) xx100`
` = 23.75 xx4`
` = 95.00 % `
Discussion
12120.

A thin rod of mass 6m length 6L is bent regular hexagon. The M.I. of the hexagon about a normal axis to its plane and through centre of system is

Answer»

`mL^(2)`
`3mL^(2)`
`5mL^(2)`
`11mL^(2)`

ANSWER :C
Discussion
12121.

If the excess pressure inside a spherical soap bubble of radius 0.1 m is balanced by that due to a column of oil of specific gravity 0.9,1.36xx10^-3 m high, calculate the surface tension.

Answer»

SOLUTION :`0.03N. M^-1`
Discussion
12122.

If the percentage increase in length of th ependulum is 3% and percentage increase in acceleration due to gravity is 1% then the percentage increase in time period of the pendulum is

Answer»

`1%`
`0.5%`
`1.5%`
0.02

Answer :A
Discussion
12123.

The moment of inertia of a disc pivoted at its centre about the axis AB as shown in the figure is I. If temperature of disc is increased by Delta T, then momentof inertial of disc about AB increases by

Answer»

`2 ALPHA I DELTA T`
`gt 2 alpha I Delta T`
`LT 2 alpha Delta T`
Can't say anything

Answer :C
Discussion
12124.

A beadof mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equations is x^(2) = 4 ay . The wire frame is fixed and the beadcan slide on it without friction . The bead is released from the point y = 4a on the wire frame from rest . The tangential acceleration of the bead when it reaches the position given by y = a is :

Answer»

`(g)/(2)`
`(sqrt3g)(2)`
`(g)/(sqrt2)`
g

Solution :`x^(2) = 4 AY , (dy)/(dx) = (1)/(2a)x`
at point (2a , 2a) , `(dy)/(dx) = 1 , theta = 45^(@)`
Component of weight ALONG tangetial direction is mg SIN `theta`.
Tangential acceleration `(dv)/(dt) = omega (DR)/(dt`
`omega = 4 PI , (dr)/(dt) = 0.5` , a = `2 pi`
`T - "mg" = "ma" , T = "mg" + "ma"`
Discussion
12125.

An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5cm and 3.75cm. The ratio of the velocities in the two pipe is

Answer»

`9:4`
`3:2`
`SQRT(3):sqrt(2)`
`sqrt(2):sqrt(3)`

ANSWER :A
Discussion
12126.

Two point masses m are connected the light rod of length l and it is free to rotate in vertical plane as shown in figure. Calculate the minimum horizontal velocity is given to mass so that it completes the circular motion in vertical lane.

Answer»

Solution :Here tenstion in the rod at the TOP most POINT of circle can be zero or negative for completing the loop. So velocity at the top most point is zero.
From energy conservation,
`1/2mv^2+1/2mv^2/4=mg(2L)+mg(4l)+0`
`v=sqrt((48gl)/(5))`
Discussion
12127.

"Water is slightly warmer at the base of a water fall than at the top". Why?

Answer»

SOLUTION :The POTENTIAL energy of the water at the TOP is CONVERTED into heat energy at the BOTTOM, which raises the temperature of water.
Discussion
12128.

One face of the a cube of side 0.2 m is in contact with ice and the opposite face is in contact with steam. If all other sides are well lagged, calculate the mass of ice that melts during one hour. Thermal conductivity of the metal =40 "Wm"^(-1) "K"^(-1). L.H. of fusion of ice =336 "kJ kg"^(-1).

Answer»


Solution :`m = KA ( theta_1 - theta_2) t// ANGLE X = 40 xx 0.04 xx 100 xx 3600//336 xx 10^(3)xx 0.2 = 8.571 "KG"`
Discussion
12129.

Explain Hydrostatic Paradox.

Answer»

Solution :Below figure isa illustration of hydrostatic paradox.

Pressure at any point in the liquid does not depend on the shape or area of vessel .This FACT is known as hydrostatic paradox.
Consider vessels of different shape asshown in figure .
They are CONNECTED at the BOTTOM by a horizontal pipe.
On FILLING with water the level in the three vessels is the same though they hold different amount of water .
This SHOWS that water at the bottom has the same pressure below each section of the vessel.
Discussion
12130.

Springs of constants k, 2k, 4k, 8k, 16k ...... are connected in series. The mass 'm' kg is attached to the lower end of the last spring and the system is allowed to vibrate. The frequency of oscillation is

Answer»

`(1)/(pi)sqrt((m)/(2k))`
`(1)/(2PI)sqrt((K)/(2m))`
`(1)/(2pi)sqrt((2m)/(2k))`
`(1)/(2pi)sqrt((4M)/(k))`

Answer :B
Discussion
12131.

The device shown in fiugre can be used to measure the pressure and volume flow rate when a person exhales. There is a slit of width b running down th length of the cylinder. Inside the tube there is a light movable piston attached to an ideal spring of force constantK. In equiilibrium position the piston is at a position wherethe slit starts (shown by line AB in the figure. A person is made to exhale into the cylinder causing the piston to compress the spring. Assume that slit width b is very small and the outflow area is much smaller tan the cros section of the tube, even at the pistons full extension. A person exhales and the spring compresses by x. (Density of air=rho) (a) Calculate the gage pressure in the tube. (b) Calculate the volume flow rate (Q) of the air.

Answer»


Answer :(a) `(KX)/(PIR^(2))` (B) `bx^(3//2)sqrt((2K)/(rhopiR^(2)))`
Discussion
12132.

The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsec?

Answer»

SOLUTION :1 LIGHT YEAR `=9.467xx10^5m` 1 PARSEC `=3.08xx10^16mtherefore4.29` light year `=(4.29xx9.467xx10^15)/(3.08xx10^16)=1.32 ` parsec
Discussion
12133.

A uniform chain of length L and mass m is lying on a smooth table. One-third of its length is hanging vertically down over the edge of the table. How much work need to be done to pull the hanging part back to the table?

Answer»

Solution :Mass of hanging part of chain `= m//3`
POSITION of centre of GRAVITY below table `= L//6`
`:.` Work done = Change in POTENTIAL energy
or Work done `= (m/3) g(L/6) = (mgL)/(18)`
or Work required to be done = `(mgL)/(18)`
Discussion
12134.

Does the principle of conservation of energy always hold in case of superposition of two sound waves ?

Answer»

SOLUTION :No, do not HOLD ALWAYS.
Discussion
12135.

Explain - product of mass and velocity is important in producing effect of force.

Answer»

SOLUTION :Whenequalforceisappliedon twoobjectofdifferentmass forequaltimeintervallighterobjectwillacquiremove velocity.butchangeinmomentum of bothobjectis EQUAL.
Discussion
12136.

S= A(1-e^(-Bxt)) where S is speed, t is time and x is displacement. Then unitof B is

Answer»

`m^(-1)s^(-1)`
`m^(-2)s`
`s^(-2)`
`s^(-1)`

Solution :The power of EXPONENT is ALWAYS dimensionless.
`:. [BXT]= [M^(0)L^(0)T^(0)], [B]=(1)/([xt])= (1)/([M^(0)LT])= [M^(0)L^(-1)T^(-1)]`
Hence UNIT of B is `m^(-1)s^(-1)`
Discussion
12137.

Assertion: Gravitational potential is a scalar quantity and its unit is J/kg. Reason: From this expression, g' = g|1 - d/(R_e)| it is explained that as we go deep into the earth, gravity decreases. i.e. As d increases, g' decreases. At the centre of earth. g' = 0.

Answer»

ASSERTION and REASON are correct and Reason is the correct explanation of Assertion
Assertion and Reason are true but Reason is the FALSE explanation of the Assertion
Assertion is true but Reason is false
Assertion is false but Reason is true

Answer :B
Discussion
12138.

(A) : The rate of change of total momentum of a many particle system is proportional to the sum of the internal forces of the system.(R ) : Internal forces can change the kinetic energy and the momentum of the system.

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 but (R ) is false
(A) is false and (R ) is false

Answer :D
Discussion
12139.

Given the vectors vec(A) = 2 hat(i) + 3hat(j) - hat(k), vec(B) = 3hat(i) - 2hat(j) - 2 hat(k) " & " vec( C) = p hat(i) + p hat(j) + 2 p hat(k) find the angle between (vec(A) - vec(B)) " & " vec( C)

Answer»

`THETA = cos^(-1) ((2)/(sqrt3))`
`theta = cos^(-1) ((sqrt3)/(2))`
`theta = cos^(-1) ((sqrt2)/(3))`
NONE of these

Answer :C
Discussion
12140.

Give three examples of floating bodies.

Answer»

SOLUTION :(i) A PERSON can swim in sea water more easily than in river water.
(ii) Ice floats on wate.
(iii) The ship is made of steel but its interior is made HOLLOW by GIVING it a concave shape.
Discussion
12141.

Two blocks A and B of masses m_1 = 3 kg and m_2 = 6 kg respectively are connected with each other by a spring of force constant k = 200 N/m as shown in figure. Blocks are pulled away from each other by x_0 = 3 cm and then released. When spring is in its natural length and blocks are moving towards each other, another block of mass m 3 kg moving with velocity v_0 =0.4 m/s (towards right) collides with A and gets stuck to it. Neglecting friction, calculate (a) Velocities v_1 and v_2 of the blocks A and B respectively just before collision and their angular frequency . (b) Velocity of centre of mass of the system, after collision (c) Amplitude of oscillations of combined body (d) Loss of energy during collision.

Answer»

SOLUTION :`(a) 0.2 m//s , 0.1 m//s , 10 ` rad/s (B) 0.1 m/s (towards right) , (c) 4.8 cm , (d) 0.03 J
Discussion
12142.

An insulated container containingmonoatomic gas of molar mass m is moving with a velocity V_(0). If the container is suddenly stopped , find the change in temperature .

Answer»

`(mv_(0)^(2))/(2R)`
`(mv_(0)^(2))/(3R)`
`(3mv_(0)^(2))/(2R)`
`(1)/(2)(mv_(0)^(2))/(R)`

Solution :Here, M= mole mass of the gas `v_(o)` = initial speed
Mass of gas = n m (n= number of moles in the gas) `therefore` Initial KINETIC energy of the gas `=1/2(nm)v^(2)` FINAL KE of gas = 0
Change in Kinetic energt, `DeltaK=1/2(nm)v^(2)`
Let change in temperature of gas = `DELTAT`
Change in internal energy of the gas
`DeltaU=nC_(v)DeltaT=n((3)/(2)R)DeltaT`
As, `DeltaU=DeltaK`
`implies3/2nRDeltaT=1/2mnv_(o)^(2)`
`impliesDeltaT=(mv_(0)^(2))/(3R)`
Discussion
12143.

Which is the largest unit out of astronomical unit, light year and Parsec ?

Answer»

Solution :`1pc=3.098x10^(16) m`
1 light YEAR `=9.46xx10^(15)m`
1 astronomical unit `=1.5xx10^(11) m`
`:.` Parsec is largest unit
Discussion
12144.

A carnot engine takes in 3 xx 10^6 cal of heat from source at 627^@ C and gives it to a sink at 27^@ C. The work done by the engine would be _____ (J=4.2 J/cal).

Answer»

zero
`4.2xx10^6` J
`8.4xx10^6` J
`16.8 xx10^6` J

Solution :Temperature of HEAT source `T_1`=273+627=900 K
Temperature of heat sink `T_2`= 273+27 = 300 K
Heat absorbs from source ,
`Q_1=3xx10^6` cal
`=4.2xx3xx10^6` Joule
`rArr` The EFFICIENCY of CARNOT engine,
`eta=1-T_2/T_1`
`=1-300/900`
`=600/900`
`=2/3`
Now efficiency of carnot engine `eta=W/Q_1`
`therefore W=eta Q_1`
`=2/3xx4.2xx3xx10^6`
`=8.4xx10^6` J
Discussion
12145.

Under the action of a force, a 2kg body moves such that its position x as a function of time t is given by x = (t^(3))/(3), x is in metre and t in second. Calculate the work done by the force in the first 2 second.

Answer»

Solution :From work-energy THEOREM, `W = Delta KE , x = t^(3)//3 THEREFORE "velocity v" = (dx)/(dt) = t^(2)`
At t = 0, `v_(i) = 0^(2) = 0` , At t = 2, `v_(f) = 2^(2)` = 4 m/s
work done `W = (1)/(2) m(v_(f)^(2)-v_(i)^(2)) = (1)/(2) XX 2(4^(2) - 0) = 16 J`
Discussion
12146.

A graph sheet divided into squares each of size 1mm^2 is kept at distance of 7cm from a magnifying glass of focal length of 8cm. The graph sheet is viewed through the magnifying lens keeping the eye close to the lens. Find (i) the magnification produced by the lens (ii) the area of each square in the image formed (iii) the magnifying power of the magnifying lens. Why is the magnification found in (i) different from the magnifying power?

Answer»

Solution :(i) `u=-7cm, f=+8cm, v=?`
For a lens `1/f=1/v=1/u`
`1/(+8)=1/v-1/(-7)=1/8-1/7=1/56, v=-56cm`
Magnification `M=v/u=(-56)/(-7)=+8`
(ii) Each square is of size `1mm^1` I.e its length and BREADTH are each to 1mm. The virtual image formed has LINEAR magnification 8. So its length and breadth are each equal to 8mm. The area of the image of each square= `8 times 8 mm^2=64mm^2`
(iii) Magnifying power of the magnifying glass i.e., simple microscope
`m=1+D/f=1+25/8=4.125 (THEREFORE D=25cm)`
the magnification FOUND in (i) is different from the magnifying power because the image DISTANCE in (i) is different from the least distance of distinct vision, D.
Discussion
12147.

What is the most likely value for C_r(molar heat capacity at constant temperature)?

Answer»

0
`0 LT C_r lt C_v`
`C_v lt C_r lt C_p`
INFINITY

ANSWER :D
Discussion
12148.

A smooth block is released from rest on a 45^(@) inclined plane and it slides a distance 'd'. The time taken to slide is n times that on a smooth inclined plane. The coefficient of friction

Answer»

`mu_(k)=1-(1)/(n^(2))`
`mu_(k)=sqrt(1-(1)/(n^(2)))`
`mu_(k)=(1)/(1-n^(2))`
`mu_(k)=sqrt((1)/(1-n^(2)))`

SOLUTION :`mu_(k)=tan THETA(1-(1)/(n^(2)))`
Discussion
12149.

A fly wheel is a uniform disc of mass 72 kg and radiu 50 cm. When it is rotating at the rate of 70 rpm, its kinetic energy is

Answer»

142 J
242 J
342 J
300 J

Solution :Kinetic energy of flywheel , `K = (1)/(2) I omega^(2) = (1)/(2) xx (1)/(2) MR^(2) omega^(2) = (1)/(4) MR^(2) omega^(2)`
=`(1)/(4) xx 72 ((50)/(100))^(2) ((2pi xx 70)/(60))^(2) = 241.5 J- 242 J`.
Discussion
12150.

A fireman wants to slide down a rope. The breaking load for the rope is 3/4th of the weight of the man. With what minimum acceleration should the fireman slide down ? Acceleration due to gravity is g.

Answer»

Zero
`(G)/(4)`
`(3g)/(4)`
`(g)/(2)`

Answer :B
Discussion