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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 15601. |
In a device designed by Academician Rebinder the surface tension is determnined from the pressure difference required to form a bubble of air at the end of a capillary immersed in the liquid being investigated (fig). Calculated the surface tension if the radius of the capillary is r = 1 mm and the difference in the pressures during bubble formation is DeltaP = 14 mm of water column. The end of the capillary is near the surface of the liquid. |
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Answer» `T = 2SR = 2 xx .030 xx (6.28 xx 10^(-2))/(2 xx 3.14) = 6 xx 10^(-4) N` 
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| 15602. |
The period of a simple pendulum on the surface of earth is T. At an atitude of half of the radius of earth from the surface, its period will be |
| Answer» Answer :B | |
| 15603. |
Using Pascal's law, explain with a neat figure, the working of a hydraulic lift. Mention any one situation from your daily life where hydrostatic paradox is useful. |
| Answer» SOLUTION :Hydrostatic PARADOX is USED to lift HEAVY VEHICLES in automobile garages and weighbridges. | |
| 15604. |
A force F acting on a body depends on its displacement s as F alpha s^(-1//3). The power delivered by F will depend on displacement as |
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Answer» `S^(2//3)` |
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| 15605. |
Consider the collision depicted in Fig. 6.10. to be between two billiard balls with equal masses m_(1)= m_(2). The first ball is called the cue while the second ball is called the target. The billiard player wants to 'sink' the target ball in a corner pocket, which is at an angle theta_(2)= 37^(@). Assume that the collision is elastic and that friction and rotational motion are not important. Obtain theta_(1). |
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Answer» Solution :From momentum conservation, since the masses are equal `v_(1i)= v_(1f)+v_(2f)` or, `v_(1i)""^(2)=(v_(1f)+v_(2f))*(v_(1f)+v_(2f))` `=v_(1i)""^(2)+v_(2f)""^(2)+2v_(1f)*v_(2f)` `={v_(1f)""^(2)+v_(2f)""^(2)+2v_(1f)v_(2f) cos(theta_1 +37^(@))}""(6.32)` Since the collision is elastic and `m_(1)= m_(2)` it follows from conservation of kinetic energy that `v_(1f)""^(2)= v_(1f)""^(2)+v_(2f)""^(2)` (6.33) Comparing Eqs. (6.32) and (6.33), we get `cos (theta_(1)+37^(@))=0` or, `theta_(1)+ 37^(@)= 90^(@)` THUS, `theta_(1)= 53^(@)`. This proves the FOLLOWING result : when two equal masses undergo a glancing elastic collision with one of them at rest, after the collision, they will MOVE at right angles to each other. |
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| 15606. |
Briefly explain the difference between travelling waves and standing waves. |
Answer» SOLUTION :
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| 15607. |
The pulley arrangements to Fig.(A) and (B), are identical. The mass of the rope is negligible. In (A) the mass mi s lifted up by attaching a mass 2m to the other end of the rope. In (B) m is lifted up by pulling the other end of the rope with a constant downward force F=2mg. In which case the acceleration of 'm' is more? |
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| 15608. |
A uniformly moving cricket balls is hit with a bat for a very short time and is turned back. Show that variation of its acceleration with time taking the acceleration in the backward direction as positive. |
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| 15609. |
Give the formula that can be used to determine refractive index of material of a prism in minimum deviation position. |
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Answer» Solution :`mu = (SIN(A + delta_m)//2)/(sin A//2)` SYMBOLS have USUAL MEANING. |
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| 15610. |
A boat crosses a river of width 200m in the shortest time and is found to experience a drift to 100m in reaching the opposite bank. The time taken now is 't' . If the same boat is to cross the river by shortest path. The time taken to cross will be |
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Answer» 2t |
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| 15611. |
The length of a seconds pendulum is decreased by 1%. What is the gain or loss in time per day? |
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Answer» Solution :`T PROP sqrtl,T_2//T_1=sqrt(l_2//l_1)T_2//2=sqrt(0.99l//l)=0.9949.T_2 =1.9898` Decrease in period = 2 - 1.9898 = 0.0102 s. GAIN in one DAY `= 0.0102 xx24xx60xx60//2 = 440.64 s`. |
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| 15612. |
He is filled in a closed vessel.When it is heated from 300K to 600K .The average K.E will be |
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Answer» Half |
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| 15613. |
A liquid cools from70^(@)C to60^(@)C in 5 minutes. If the temperautre of the surrounding is 30^(@)C, what is the time taken by the liquid to cool from 50^(@)C to 40^(@)C ? |
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Answer» Solution :According to Newton.s law of cooling, Rate of COOLONG`prop` = {MEAN difference of twemperature between the body and the surroundings `(theta_(2)-theta_(1))/t prop((theta_(1)+theta_(2))/2)-theta_(0)` where t is the time TAKEN by the liquid to cool from `theta_(2)^(@)`C to `theta_(1)^(@)`C and `theta_(0)` is the TEMPERATURE of the surrounding `(60-50)/5 prop((60+50)/2)-30` `2 prop 55-30` `2 prop 25` ...(i) Let the time taken to cool from `50^(@)`C to `40^(@)`C be t, `(50-40)/t prop ((50+40)/2)-30` `10/(t) prop15` ...(ii) Dividing EQUATION (i) by (ii) `(2xxt)/10 = 25/(15) , t = (25xx10)/(2xx15)` = 8.33 s |
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| 15614. |
What do you mean by the term weightlessness? |
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Answer» Solution :`(i)` Freely falling objects experience only gravitational force. As they fall freely , they are not in contact with any surface (by neglecting air friction). The NORMAL force acting on the object is zero. `(ii)` The downward acceleration is EQUAL to the acceleration DUE to the gravity of the EARTH. i.e. `(a=g)` `(iii) a=g :. N=m(g-g)=0`. `(iv)` This is called the state of weightlessness. `(v)` EG. When the lift falls (when the lift wire cuts) with downward acceleration `a=g`, the person inside the elevator is in the state of weightlessness. |
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| 15615. |
Which one of the following is not a periodic motion? |
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Answer» Rotation of the earth about its axis. |
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| 15616. |
(A) : When couple acting on a body moment of ht couple does not depends on the point about which momentsare measured.(R ) : When a couple acts on a body it is in transtational equilibrium. |
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Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) |
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| 15617. |
One end of a string is tied to a rigid support attached to roof and the other end is wound around a uniform wheel of radius r and mass m. The string is kept vertical and the wheel is released from rest. As the wheel descends show that (i) the tension in the string is one-third the weight of the wheel, (ii) the magnitude of the acceleration of the centre of mass is 2g/3 and (iii) the speed of the centre of mass is (4gh//3)^(1//2) |
| Answer» SOLUTION :`mg -T = ma, tau = IALPHA, Tr =(1//2) mr^(2)a, a=ralpha, T=(1//2) ma, (3//2) ma = mg, a =2g//3, MGH = (1//2) mv^(2) + (1//2)(1//2)mr^(2)(v^(2)//r^(2)), v=(4gh//3)^(1//2)` | |
| 15618. |
For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase? |
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Answer» Solution :All the WAVES have different phases. The given transverse harmonic wave is: `y(x,t)=3.0 "sin"(36+ 0.018x+(pi)/(4))""…(i)` For x = 0, the equation reduces to: `y(0,t)=3.0 "sin"(36t+(pi)/(4))` Also, `omega=(2pi)/(T)=36 rad//s^(-1)` `therefore T=(pi)/(8) s` Now, plotting y vs. t graphs using the different values of t, as listed in the given table.  For x = 0, x = 2, and x = 4, the phases of the THREE waves will get changed. This is because amplitude and frequency are INVARIANT for any CHANGE in x. The y–t PLOTS of the three waves are shown in the given figure. 
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| 15619. |
A particle is projected from point P with velocity 5sqrt(2)ms^(-1) perpendicular to the surface of a hollow right angle cone whose aaxis is vertical. It collides at Q normally. The time of the flight of the particle is |
Answer» SOLUTION :`t=u/(G SIN THETA)=(5sqrt(2)xxsqrt(2))/10=1sec` 
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| 15620. |
(A): The ratio of rms speed and average speed of gas molecules at a given temperature istemperature is sqrt(3):sqrt((8)/(pi)) (R ):C_(rms)gtC_(av) |
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Answer» Both (A) and ( R) are true and (R ) is the correct EXPLANATION of (A) |
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| 15621. |
When white light enters a prism. It gets split into its constituent colours. This is due to |
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Answer» HIGH DENSITY of prism material |
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| 15622. |
In which mode of vibration does the density of the medium remains the same ? |
| Answer» Solution :The DENSITY of medium remains the same in the TRANSVERSE MODE of VIBRATION. | |
| 15623. |
(a)Prove that bodies starting at the same time t = 0 from the same point, and following frictionless slopes in different directions in the samevertical plane, all lie in a circle at any subsequent time. (b) Using the above result do the following problem. A point P lies above an inclined plane of inclination angle alpha. P is joined to the plane at number of points by smooth wires, running in all possible directions. Small bodies (in shape of beads) are released from P along all the wires simultaneously. Which body will take least time to reach the plane. |
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| 15624. |
The maximum displacement of an oscillating particle is 0.05m. If its time period is 1.57s (i) What is the velocity at the mean position ? (ii) What is its acceleration at the extreme position ? |
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Answer» `0.2ms^(-1), 0.2ms^(-2)` |
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| 15625. |
The total radiation emitted by a perfectly black body is proportional to |
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Answer» the ABSOLUTE temperature |
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| 15626. |
A body constrained to move in y-direction is subjected to a force given by vecF=(-2hat i +15hatj +6hatk) N. The work done by this force in moving the body a distance of 10 m along the y-axis is |
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Answer» 20 J `VECS = 0hati + 10 hatj + 0hatk m` `:.` Work done = `vec(F).vec(S)` `= (-2hati + 15 hatj + 6 hatk) . (0hati + 10 hatj +0 hatk) = 150 J`. |
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| 15627. |
The potential energy (in SI units) of a particle of mass 2 kg in a conservative fields is U = 6x - 8y. If the initial velocity of the particle is vec(u) = - 1.5 hat(i) + 2 hat(j), then find the total distance travelled by the particle in the first two seconds. |
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| 15628. |
A ship is moving at a constant speed of 10 km//hr in the direction of the unit vector hat(i). Initially. Its position vector, relative to a fixed origin is 10(-hat(i)+hat(j)) where hat(i) & hat(j) are perpendicular vectors of length 1 km. Find its position vector relative to the origin at time t hours later. A second ship is moving with constant speed u km//hr parallele to the vector hat(i)+2hat(j) and is initially at the origin (i) If u=10 sqrt(5) km//h. Find the minimum distance between the ships and the corresponding value of t (ii) Find the value of u for which the shipd are on a collision course and determine the value of t at which the collision would occur if no avoiding action were taken. |
Answer»  `:.` DISPLACEMENT after TIME `'t' = 10 hat (i) xx t` Velocity of second ship `= u xx ((hat (i) +2 hat(j)))/(sqrt(5))` `tan theta = (2u//5)/((10 -(u)/(sqrt(5))))=(2 xx 10 sqrt(5))/(10 sqrt(5) -10 sqrt(5))` (i) `t = (10)/(20) = (1)/(20) sec`, minimum distance = 10 KM. |
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| 15629. |
If the direction of torque is out of the paper then the rotation produced by the torque is |
| Answer» Answer :A | |
| 15630. |
Graph of a body is shown below. It explains that |
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Answer» at B FORCE is zero |
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| 15631. |
An electric pump on the ground floor of a building taken 15 minutes to fill a tank of volume 30 m^(3) with water. If the tank is 40 m above the ground and the efficiency of the pump is 30%, find the electric power consumed by the pump in filling the tank. [density of water = 10^(3)kg//m^(3)] |
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Answer» SOLUTION :Mass of the water to be filled = (volume of the tank) `XX` density of water `m=30xx10^(3)KG` `h=40 m, t=15xx60=900` sec. Efficiency `eta = 30%` Inputpower `P_(i)=(100)/(eta)xx(mgh)/(t)` `= (100xx30xx10^(3)xx9.8xx40)/(30xx900)=43.56 KW` |
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| 15632. |
If a body is projected with speed less than escape velocity, then a) the body can reach a certain height and may fall down following straight line path b) the body can reach a certain height and may fall down following a parabolic path c) the body may orbit the earth in a circular path d) the body may orbit the earth in an elliptical path. |
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Answer» only a & B are TRUE |
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| 15633. |
When we try to close a water tap with our fingers fast jets of water through the openings between our fingers. Explain why. |
| Answer» SOLUTION :From the equation of continuity, `va=` constant where, v is the velocity of the fluid and a is cross sectional area of the tube. When we TRY to close a water tap with our fingers , the cross sectional area of the outlet of water jet is reduced . HENCE , velocity of water INCREASES greatly and FAST jets of water come through the openings between our fingers. | |
| 15634. |
A piece of stone came to rest after covering a distance of 20 m over an surface. If the initial velocity of the stone was 1 "m.s"^(-1), what was the coefficient of friction between the stone and ice? |
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Answer» |
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| 15635. |
If the change in the value of g at a height h above the surface of the earth is same as at depth .x. below the surface of the earth, then (h |
| Answer» ANSWER :B | |
| 15636. |
A pendulum oflength 1 m and having a bob of mass 1 g is pulled aside through an angle 60^(@) with the verical and then released. The power delivered by all the forces acting on the bob when the pendulum makes 30^(@) with the vertical is ______ (g = 10 ms^(-2)) |
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Answer» 13.5 mW |
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| 15637. |
Find the fractionalincreasein volumeof a wireof circularcross section if its longitudinalstrainis 1% . (sigma=0.30) |
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Answer» SOLUTION :we know that `(dV)/(V) =(dL)/(L) (1-2sigma)` Here, `(dL)/(L) =1% =(1)/(100) =0.01, SIGMA =0.30`. The FRACTIONAL increase in VOLUME, `(DeltaV)/(V) =0.01 (1-2xx0.30) =0.01 xx 0.40 =0.004`. |
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| 15638. |
The level of water in a vesel falls from a height 64 cm to 36 cm in 5 minutes as water flows out through a hole inits bottom. What further time is required for the water level to fall from 36 cm to 16 cm? |
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Answer» 5 minutes |
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| 15639. |
If vecu = 2vec(i)-2vec(j)+3vec(k) and the final velocity is vecv = 2vec(i) - 4hat(j) + 5hat(k) and it is covered in a time of 10 sec, find the acceleration vector. |
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Answer» `(3vec(i) - 2vec(J) + 2vec(k))/(10)` |
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| 15640. |
A spaceship is moving with an acceleration of 9.8 m cdot s^(-2) in a fixed direction in space, where attraction due to other planets and stars is negligible. Find the magnitude and direction of the force exerted by a body of 98 N, in the spaceship. |
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| 15641. |
A vessel filled with air under pressure p_(0) contains a soap bubble of diameter d. The pressure having been reduced isothermally n fold, the bubble diameter increased eta fold. Calculate the surface tension of siap water solution. |
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Answer» <P> |
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| 15642. |
What is a pure spectrum ? |
| Answer» SOLUTION :A PURE SPECTRUM is that in which there is no OVERLAPPING of COLOURS. | |
| 15643. |
If a black body is radiating at T = 1650K, at what wavelength is the intensity maximum ? |
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Answer» Solution :According to Wien.s LAW `lamda_(m)T=2.9xx10^(-3)MK` Substituting the value of the TEMPERATURE T = 1650K `lamda_(m)=(2.9xx10^(-3))/1650=1.8mum` |
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| 15644. |
SI unit and CGS unit of certain quantity vary by 10^3 times. That quantity is |
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Answer» BOLTZMANN constant |
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| 15645. |
A gun can fire shells with maximum speedv_0 and the maximum horizontal range that can be achieved isR = v_0^2 //g If a target farther away by distanceDelta x (beyound R) has to be hit with the asme gun 9Fig. 2 (EP).24). Show that it could be accieved by raising the gun to a height at least h= Delta x [ 1 + (Delta x0/R]. . |
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Answer» Solution :As per wuestion, the maximu , bborizontal range on ground is ` R_(max) = (v_0&2 /g` It isif ` theta = 54^@` . The shell fired at (0) from HEIGHT (h) can hit the same target at B. Fig. 2 ( EF). 25 ` Taking vertical downward motion from (O) to (B), we have ` ` u_y v_0 SIN theta , a_y = g, y_0 =0 , y=h, t=?` as, ` y= y_0 + u_y + 1/2 a_y t^2` :. ` x=x _0 + u_x + 1/2t+ a_x t^2` (R+ Delta x) =0 + v) cos thea t` or ` t= ((R + delta x))./(v_0 cos theta)` PUTTING value fo (t) in (i), we have ` h=- v_0 sin theta xx (R + Delta x)/(v_0cos theta) + 1/2 g ((R + Delta x^2)/(v_0^2 cos theat ) =- (R + Delat x) TAN theta + 1/2 g ((R + Detal x)^2)/(v_0^2 cos theta^2 theta)` As ` theta = 45^@` , so ` h=- (R + Deltax) tan 45^@ + 1/2 (g(R+ Delta x)^2)/( 2 v_0^2 cos^@ 45^@` or ` h=- (R + Delta x) xx 1 + 1/2 (g(R+ Delta x)^2)/v_0^2` xx 2 =- (R + Delta x) + 1/ R xx (R + Delta x)^2` [:. v_0^2 //g = R]` ` =- (R + Delta x) + R + 2 Delta x + Delta x^2 //R =Delta x + Delta x^2 //R = Delta x (1 + Delta x//R)`. 
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| 15646. |
If a car covers 2//^(th) of the total distance with v_(1)speed and 3//5^(th) distance with v_(2) then average speed is |
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Answer» `1/2 SQRT(v_(1)v_(2))` |
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| 15647. |
Two piston of a hydraulic lift have diameters of 60 cm and 5 cm. What is the force exerted by the larger piston when 50 N is placed on the smaller piston ? |
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Answer» Solution :SINCE, the diameter of the pistons are given, we can calculate the RADIUS of the piston `r = D/2` Area of SMALLER piston, `A_1 = pi(5/2)^(2) = pi(2.5)^(2)` Area of larger piston, `A_2 = pi(60/2)^2 = pi(30)^(2)` `F_2 = (A_2)/(A_1) xx F_1 = (50 N) xx (30/2.5)^2 = 7200 N` This means, with the force of 50 N, the force of 7200 N, can be LIFTED. |
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| 15648. |
If C is capacitance, Vis potential, rho is specific resistance and epsilon_(0) is permittivity of free space, then the dimensions of (CV)/(rhoepsilon_(0))are same as that of |
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Answer» charge `[RHO]= [ML^(3)T^(-3)A^(-2)], [epsilon_(0)]= [M^(-1)L^(-3)T^(4)A^(2)]` `:.[(CV)/(rhoepsilon_(0))]= ([C][V])/([rho][epsilon_(0)])= ([M^(-1)L^(-2)T^(4)A^(2)][ML^(2)T^(-3)A^(-1)])/([ML^(3)T^(-3)A^(-2)][M^(-1)L^(-3)T^(4)A^(2)])= ([AT])/([T])= [A]= ["Current"]` |
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| 15649. |
Five kilomoles of oxygen is heated at constantpressure. The temperature of the oxygen gas is increased from 295K to 305K. If the molar heat capacity of oxygen at constant pressure is 6.994 kcal/kmole K. The amount of heat absorbed is in kcal, |
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Answer» 249.7 |
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| 15650. |
The least coefficient of friction for an inclined plane inclined at an angle alpha with horizontal, in order that a solid cylinder will roll down it without slipping ? |
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Answer» ⅔ `TAN ALPHA` |
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