1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 3 InterviewSolution

101.	An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring isA. `4 Mg//k`B. `2Mg//k`C. `Mg//k`D. `Mg//2k`
Answer» Correct Answer - B

Discussion

102.	A block of mass `2kg` is propped from a heught of `40cm` on a spring where force constant is `1960Nm^(-1)` The maximum distance thought which the spring compressed byA. 5 cmB. 15 cmC. 20 cmD. 10 cm
Answer» Correct Answer - D `mg(h+x)=1/2kx^2` Here, m= 2 kg, h=40 cm `k=1960 N m^(-1) and g=10 m s^(-2)` `therefore 2 xx 10 (0.40+x)=1/2xx1960 x^2` On solving , we get x = 10 cm

Discussion

103.	A block of mass 1 kg is attached to one end of a spring of force constant k = 20 N/m. The other end of the spring is attached to a fixed rigid support. This spring block system is made to oscillate on a rough horizontal surface `(mu = 0.04)`. The initial displacement of the block from the equilibrium position is a = 30 cm. How many times the block passes from the mean position before coming to rest ? `(g = 10 m//s^(2))`A. 11B. 7C. 6D. 15
Answer» Correct Answer - B

Discussion

104.	A block of mass 0.1 kg attached to a spring of spring constant 400 N/m is putted rightward from `x_(0)=0` to `x_(1)=15` mm. Find the work done by spring force.
Answer» `S_(s)=-1/2kx^(2)` `W_(s)=-1/2kx^(2)` `=-1/2xx400xx(15x10^(-3))^(2)` `=-0.045J`

Discussion

105.	Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetricaly tostretch the spring by a length x over its natural length. The work doen by the spring oneach mass isA. `(1)/(2)kx^2`B. `-(1)/(2)kx^2`C. `(1)/(4)kx^2`D. `-(1)/(4)kx^2`
Answer» Correct Answer - D `triangleU=(1)/(2)kx^2` Work done by spring on each mass `=-(1)/(4)kx^2`

Discussion

106.	One end of a spring is rigidly fixed. A block attached to the free end of the spring is pulled through a distance `x_(0)`. On releasing the block, its amplitude of motion cannot exceed `+-x_(0)`. Why?
Answer» When the spring is stretched by `x_(0)` its P.E. `=(1)/(2)kx_(0)^(2)` where k is spring constant. At `xgtx_(0)`. PE will be greater than `(1)/(2)kx_(0)^(2).` Therefore , its `KE=((1)/(2)kx_(0)^(2)-(1)/(2)kx^(2))` would be negative, which is not possible. Therefore, amplitude of motino cannot exceed `+-x_(0)`.

Discussion

107.	Two equal masses are attached to the two ends of a spring of spring constant `k`. The masses are pulled a part symmetrically to stretch the spring by a length `x` over its natural length. The work done by the spring on each mass is.A. `1/2 kx^(2)`B. `-1/2 kx^(2)`C. `1/4 kx^(2)`D. `-1/4 kx^(2)`
Answer» Correct Answer - D

Discussion

108.	A ball of mass m is thrown in air with speed `v_(1)` from a height `h_(1)` and it is at a height `h_(2)(gth_(1))` when its speed becomes `v_(2)` . Find the work done on the ball the air resistance.
Answer» Work done on the ball by gravity is `W_g=-mg(h_2-h_1)` Let work done on the ball by air resistance be `W_(air)` `:. W_g+W_(air)=DeltaKE` `implies-mg(h_2-h_1)+W_(air)=1/2m(v_2^2-v_1^2)` `implies W_(air)=mg(h_2-h_1)+1/2m(v_2^2-v_1^2)`

Discussion

109.	A man is riding on a cycle with velocity `7.2(km)/(hr)` up a hill having a slope 1 in 20. The total mass of the man and cycle is 100kg. The power of the man isA. 200 WB. 175 WC. 125 WD. 98W
Answer» Correct Answer - D

Discussion

110.	A block of mass `m` at rest is acted upon by a force `F` for a time t. The kinetic energy of block after time t isA. `(Ft^2)/(m)`B. `(F^2t^2)/(2m)`C. `(Ft^2)/(2m)`D. `(F^2t^2)/(3m)`
Answer» Correct Answer - B `a=(F)/(m)` `s=(1)/(2)at^2=(Ft^2)/(2m)` `W=triangleK` `Fs=K-0impliesK=(F^2t^2)/(2m)`

Discussion

111.	A particle at rest on a frictionless table is acted upon by a horizontal force which is constant in magnitude and direction. A graph is plotted of the work done on the particle W, against the speed of the particle v. If there are no frictional forces acting on the particle, the graph will look likeA. B. C. D.
Answer» Correct Answer - D (d) Form work-energy theorem W=change in kinetic energy or`" " W=(1)/(2)mv^(2)` `:.` W-v graph is a parabola.

Discussion

112.	How much work is done by a coolie walking on a horizontal platform with a load on his head ? Explain.
Answer» W = 0 as his displacement is along the horizontal direction and in order to balance the load on his head, he applies a force on it in the upward direction equal to its weight. Thus angle between force and displacement is zero.

Discussion

113.	`300 J`of work is done in slide a `2kg` block up an inclined plane of height `10m`. `Taking g = 10 m//s^(2), work done against friction isA. `200 J`B. `100 J`C. zeroD. `1000 J`
Answer» Net Net work done in sliding a body up to a height `h` on inclined plane `= `Work done against gravitatioinal force + Work done against frictional force ` implies W = W_(g) + W_(f)` (i) but `W = 300J` `W_(g) + mgh = 2 xx 10 xx 10 = 200 J` Putting in Eq (i) we get `300 = 200 + W_(f)` `W_(f) = 300 - 200 = 100J`

Discussion

114.	300J of work is done in slinding a 2 kg block up an inclined plane of height 10m. Taking g =`10m//s^(2)` , work done against friction isA. 200JB. 100JC. zeroD. 1000J
Answer» Correct Answer - b

Discussion

115.	A shell acquires the initial velocity `v=320m//s`, having made `n=2.0` turns inside the barrel whose length is equal to `l=2.0m`. Assuming that the shell moves inside the barrel with a uniform acceleration, find the angular velocity of its axial rotation at the moment when the shell escapes the barrel.
Answer» Correct Answer - A::B::C

Discussion

116.	In the figure show `m=1kg, ` M = 2 kg. When smaller block (m) descends through 1m, it acquires a speed of 1 m/s. Take `g=m//s^(2)` and choose the correct statements out of the following A. When the speed of smaller block (m) is 1 m/s.B. When the speed of smaller block (m) is 1 m/s, the speed of bigger block is 2 m/sC. Coefficient of kinetic friction between the bigger block (M) and horizontal surface is `13/40`D. Coefficient of kinetic friction between the bigger block (M) and horizontal surface is `11/80`
Answer» Correct Answer - B::D

Discussion

117.	A `1-KW` motor pumps out water from a well `10m` deep. Calculate the quantity of water pumped out per second.
Answer» Power, `P=1kW=10^3W`, `S=10m`, time, `t=1s`, mass of water, `m=?` `Power=(mgxxS)/(t)` `10^3=(mxx10xx10)/(1)` or `m=(10^3)/(10xx10)kg=10kg`

Discussion

118.	(a) What is the power of an engine which can lift 600 kg of coal per minut from a mine 20 m deep ? (b) A 2kW motor pumps out water from a well `10m` deep. Calculated the quantity of water pumped out per second.
Answer» `overlineP=(W)/(t)=(mgh)/(t)=(600xx10xx20)/(60)=2000W` (b) `overlineP=(W)/(t)=(mgh)/(t)` `2000=(m)/(t)xx10xx10` `(m)/(t)=20` Quantity of water pumped out per second `=20kg`

Discussion

119.	Calculate the power of a crane in watts, which lifts a mass of `100 kg` to a height of `10 m` in `20 s`.
Answer» Given, mass=m=100kg ltbRgt height=h=10m time duration t=20s Power=Rate of work done `("Change in PE")/("time")=(mgh)/(t)` `(100xx9.8xx10)/(20)`=`5xx98=490W`

Discussion

120.	A crane lifts a mass of 100 kg to a height of 10 m in 20 s. The power of the crane is (Take `g = 10 m s^(-2)`)A. 100 WB. 200 WC. 250 WD. 500 W
Answer» Correct Answer - D Here, m= 100 kg, h = 10 m, t=20 s, `g= 10 m s^(-2)` The power of the crane is `P= (mgh)/t= ((100 kg)(10ms^(-2)) (10m))/((20 s))= 500W`

Discussion

121.	A crane lifts a mass of `100 kg` to a height of `10m` in `20s`. The power of the crane is `(Take g = 10 ms^(-2))`
Answer» Here, `m=100kg, h=10m, 1=20s` `P=(W)/(t)=(mgh)/(t)=(100xx9.8xx10)/(20)=490W`

Discussion

122.	Initially the system shown in figure is in equilibrium. At the moment, the string is cut the downward of `a(1)` and `a_(2)` are. .A. (a) zero and zeroB. (b) 2g and zeroC. (c) g and zeroD. (d) None of the above
Answer» Correct Answer - B `T =2mg` As soon as string is cut T (on A) suddenly becomes zero . Therefore a force of 2mg on upward direction on A suddenly becomes zero. So net force on it will become 2mg downwards. `:. a^(1) = (2mg)/(m) =2g` (downwards) Spring force dorce not become instantly zero so acceleration of B will not change abruptly. or `a_(2) =0`

Discussion

123.	A spring of force constant `k` is cut into there equal part what is force constant of each part ?
Answer» Correct Answer - C KEVCONCEPT For a (spring constant) = Constant if length is made one third , the spring , constant became three lines .

Discussion

124.	A spring of force constant k is cut in two parts at its one-third lingth. When both the parts are stretched by same amount. The word done in the two parts will be .A. equal in bothB. greater for the longer partC. greater for the shorter for the shorter partD. data insufficient.
Answer» Correct Answer - C `kprop1/l` l of shorter part is less, therefore value of k is more, `W =(1)/(2)kx^(2)` `:. W^("shorter part")` will be more.

Discussion

125.	A particle moving in a straight line is acted upon by a force which works at a constant rate and chages ist melocity from (u and v ) over a distance x. Prove that the taken in it is `3/2 (u+v)x/(u^(2)+v^(2)+uv)` .
Answer» Correct Answer - A From work-energy principle, `W=DeltaKE` `:. Pt=1/2m(v^(2)-u^(2)) (P="power")` or `t=m/(2P)(v^(2)-u^(2))` Further `F.v=P` `:. m.(dv)/(ds).v^(2)=P` or `int_(u)^(v)v^(2)dv =P/mint_(0)^(x)` `:. (v^(3)-u^(3) =(3P)/m.x` or `m/P =(3x)/(v^(3)-u^(3)))` Substituting in Eq. (i) `t=(3x(u+v))/(2(u^(2)+v^(2)+uv))` Hence proved.

Discussion

126.	The system is released from rest with the spring initally stretched `75 mm`. Calculate the velocity `v` of the block afrer it has dropped `12 mm`. The spring has a stiffness of `1050 N//m`. Neglect the mass of the small pulley.
Answer» Correct Answer - A::C If block drops 12 mm, then sping will further stretch by 24 mm. Now, `E_(i) =E_(f)` `:. 1/2 xx 1050 xx (0.075)^(2) =-45 xx 10 xx 0.012` `+(1)/(2) xx 45 xx v^(2) + 1/2 xx 1050 xx (0.099)^(2)` Solving we get, `v=0.37 m//s`.

Discussion

127.	A small object is sliding on a smooth horizontal floor along a vertical wall. The wall makes a smooth turn by an angle `theta_(0)` . Coefficient of friction between the wall and the block is `mu`. Speed of the object before the turn is u. Find its speed (V) just after completing the turn. Does your answer depend on shape of the curve? [The turn is smooth and there are no sharp corners.]
Answer» Correct Answer - `V=ue^(-mu theta_(0))`; No the answer does not depend on the shape of the curve.

Discussion

128.	A pillar having square cross section of side length L is fixed on a smooth floor. A particle of mass m is connected to a corner A of the pillar using an inextensible string of length 3.5 L. With the string just taut along the line BA, the particle is given a velocity v perpendicular to the string. The particle slides on the smooth floor and the string wraps around the pillar. (a) Find the time in which the particle will hit the pillar. (b) Find the tension in the string just before the particle hits the pillar. Neglect any energy loss of the particle.
Answer» Correct Answer - (a) `t=(4piL)/v` `(b) T= (2mv^(2))/L`

Discussion

129.	(i) A simple pendulum consist of a small bob of mass m tied to a string of length L. Show that the total energy of oscillation of the pendulum is `E~=1/2 mg L theta_(0)^(2)` when it is oscillating with a small angular amplitude `theta_(0)`. Assume the gravitational potential energy to be zero of the lowest position of the bob. (ii) Three identical pendulums A, B and C are suspended from the ceiling of a room. They are swinging in semicircular arcs in vertical planes. The string of pendulum A snaps when it is vertical and it was found that the bob fell on the floor with speed `V_1`. The string of B breaks when it makes an angle of 30° to the vertical and the bob hits the floor with speed `V_2`. The string of pendulum C was cut when it was horizontal and the bob falls to the floor hitting it with a speed V3. Which is greatest and which is smallest among `V_1,V_2` and `V_3`?
Answer» Correct Answer - (i) `V_(1)=V_(2)=V_(3)`

Discussion

130.	A particle of mass m is attached to an end of a light rigid rod of length a. The other end of the rod is fixed, so that the rod can rotate freely in vertical plane about its fixed end. The mass m is given a horizontal velocity u at the lowest point. (a) Prove that when the radius to the mass makes an angle `theta` with the upward vertical the horizontal component of the acceleration of the mass (measured in direction of u) is `[g(2+3 cos theta)-u^(2)//a] sin theta` (b) If 4ag lt`u^(2)` lt5ag, show that there are four points at which horizontal component of acceleration is zero. locate the points.
Answer» Correct Answer - The four points are represented by `theta=-0, pi cos^(-1)((u^(2)-2ag)/(3ag))` and `[2pi-cos^(-1)((u^(2)-2ag)/(3ag))]`

Discussion

131.	AB is a mass less rigid rod of length 2l. It is free to rotate in vertical plane about a horizontal axis passing through its end A. Equal point masses (m each) are stuck at the centre C and end Bof the rod. The rod is released from horizontal position. Write the tension in the rod when it becomes vertical.
Answer» Correct Answer - Tension in segment `AC =28/5 mg` Tension in Segments `BC=17/5 mg`

Discussion

132.	A light rigid rod has a bob of mass m attached to one of its end. The other end of the rod is pivoted so that the entire assembly can rotate freely in a vertical plane. Initially, the rod is held vertical as shown in the figure. From this position it is allowed to fall. (a) When the rod has rotated through `theta=30^(@)` , what kind of force does it experience– compression or tension? (b) At what value of `theta` the compression (or tension) in the rod changes to tension (or compression)?
Answer» Correct Answer - (a) Compression (b) `theta =cos^(-1)(2/3)`

Discussion

133.	Figure shows as rod of length 20 cm pivoted near an end and which is made to rotate in a horizontal plane with a constant angular speed. A ball ofmass m is spuspended by a string angular also of length 20 cm from the other end of the rod. If the angle `theta` made by the stirng with the vertical is `30^0`. find the anglular speed of the rotation. Take `g=10 m/s^2`
Answer» Correct Answer - A::D

Discussion

134.	With what minimum speed `v` must a small ball should be pushed inside a smooth vertical tube from a height `h` so that it may reach the top of the tube? Radius of the tube is `R` . A. `sqrt(2g(h + 2R))`B. `(5)/(2)R`C. `sqrt(g(5R - 2h))`D. `sqrt(2g(2R - h))`
Answer» Correct Answer - D Velocity of ball to just reach the top of the tube should be given by: `0 = v^(2) - 2 gh_(0)` Here `h_(0) = (R - h)` and in critical case velocity will be zero at topmost point. Thus `v = sqrt(2g(2R - h)`

Discussion

135.	A particle of mass `m` is released from a height `H` on a smooth curved surface which ends into a vertical loop of radius `R`, as shown. Choose the correct alernative(s) if `H=2R`. A. `H =(3R)/(2)`B. `H = 5R`C. `H =(5R)/(2)`D. None of these
Answer» Correct Answer - A Force on body at `A = sqrt2 mg` `implies F = sqrt(F_(g)^(2)) + F_(c)^(2)) implies F_( c) = mg = mv^(2)//R` `implies v^(2) = gR` Using energy conserrvation `mgH = mv^(2)//2 + mgR = 5mgR//2` `H = 3R//2`

Discussion

136.	In the figure shown, there is a smooth tube of radius `R`, fixed in the vertical plane. A ball `B` of mass `m` is released from the top of the tube. `B` slides down due to gravity and compresses the spring is fixed and end `A` is free., Initially, line `OA` makes an angle `60^@` with `OC` and finally it makes an angle of `30^@` after compression. Find the spring constant of the spring. A. `(12 mg (2 + sqrt3))/(pi^(2) R)`B. `(36 mg (2 + sqrt3))/(pi^(2) R)`C. `(18 mg)/(pi^(2) R)`D. None of these
Answer» Correct Answer - D By energy conservation `mgR (1 + cos 30^(@)) = (1)/(2) k((pi R)/(6))^(2)` `implies k = (36 mg (2 + sqrt(3)))/(pi^(2)R)`

Discussion

137.	The potential energy of a system increased if work is doneA. by the system against a conservative forceB. by the system against a non conservative forceC. upon the system by a conservative forceD. upon the system by a non conservative force
Answer» Correct Answer - a

Discussion

138.	An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of `2 ms^(1)`. The mass per unit length of water in the pipe is `100 kgm^(-1)`. What is the power of the engine?A. 400 WB. 200WC. 100WD. 800W
Answer» Correct Answer - d

Discussion

139.	force `F` on a partical moving in a straight line veries with distance `d` as shown in the figure. The work done on the partical during its displacement of `12 m` is A. 21JB. 26JC. 13JD. 18J
Answer» Correct Answer - c

Discussion

140.	The kinetic energy of a body moving along a straight line varies with time as shown in figure. The force acting on the body: A. zeroB. constantC. directly proportional to velocityD. inversely proportional to velocity
Answer» Correct Answer - B `(1)/(2) mv^(2) prop t` `(1)/(2) mv^(2) = At` where `A` is constant `v = sqrt((2A)/(m)) t^(1//2)` `v prop = sqrtt` `a = (dv)/(dt) = sqrt((2 A)/(m) (1)/(2 sqrtt))` `F = ma = sqrt(2 Am). (1)/(2sqrtt)` `F prop (1)/(v)`

Discussion

141.	The potential energy of a partical veries with distance `x` as shown in the graph. The force acting on the partical is zero atA. CB. BC. B and CD. A and D
Answer» `F = (-dU)/(dx)` it is clear that slop of `U - x` curve is zero at point `B and C`

Discussion

142.	force `F` on a partical moving in a straight line veries with distance `d` as shown in the figure. The work done on the partical during its displacement of `12 m` is A. `21 J`B. `26 J`C. `13 J`D. `18 J`
Answer» Work done = Area under `(F-x)` graph `= 2 xx (7 - 3) + (1)/(2) xx 2 xx(12 - 7)` `= 8 + (1)/(2) xx 10 = 8 + 5 = 13 J`

Discussion

143.	Velocity-time graph of a particle of mass (2 kg) moving in a straight line is as shown in Fig. 9.20. Find the word done by all the forces acting on the particle. .A. `400 J`B. `-400 J`C. `-200 J`D. `200 J`
Answer» Correct Answer - C Initial velocity of the partical : `v_(1) = 2 m//s` Final velocity of the partical . `v_(f) = 0` From work energy theorm `W_("net") = Delta K.E = K_(f) - K_(i)` `(1)/(2) m (v_(f)^(2) - v_(i)^(2)) = (1)/(2) (2) (0 - 400) = - 400 J`

Discussion

144.	The displacement of a body of mass `2 kg` veries with time `t` as `S = t^(2) + 2t`, where `S` is in seconds. The work done by all the forces acting on the body during the time interval `t = 2s` to `t = 4s` isA. `36 J`B. `64 J`C. `100 J`D. `120 J`
Answer» Correct Answer - B `v = (dx)/(dt) = 2t + 2` From work energy theorm, ` W_("net") = Delta K. E. = K_(f) - K_(i)` `= (1)/(2) m (v_(f)^(2) - v_(i)^(2))` `= (1)/(2) xx 2 [(2xx 4 + 2)^(2) - (2 xx 2 + 2)^(2)] = 64 J`

Discussion

145.	Acceleration of a particle in x-y plane varies with time as `a=(2t hati+3t^2 hatj) m//s^2` At time `t =0,` velocity of particle is `2 m//s` along positive x direction and particle starts from origin. Find velocity and coordinates of particle at `t=1s.`A. `(2t^(2) + 3t^(3)) W`B. `(2t^(2) + 4t^(4)) W`C. `(2t^(3) + 3t^(4)) W`D. `(2t^(3) + 3t^(5)) W`
Answer» `vecF = 2t hati + 3t^(2) hatj` `m (d vecv)/(dt) = 2t hati + 3t^(2) hatj (m = 1 kg)` `rArr underset(0) overset(vecv)intdvecv=underset(0)overset(t)int2thati+3t^(2) hat j " "(m=1kg)` Power `= vecF. Vecv = (2t^(3) + 3t^(5)) W`

Discussion

146.	Given that the position of the body in `m` is a function of time as follows : `x = 2t^(4) + 5t + 4` The mass of the body is `2 kg` . What is the increase in its kinetic energy onesecond after the stert of motion?A. `168 J`B. `169 J`C. `32 J`D. `144 J`
Answer» `x = 2t ^(4) + 5t + 4 implies v = (dx)/(dt) = 8 t + 5` at `t = 0, v = 5 m//s` at` t = 1 s`, `v = 8 xx 1 + 5 = 13 m//s` Increases in `KE = (1)/(2) m[13^(2) - 5^(2)] = 144 J`

Discussion

147.	The potential energy `U` in joule of a particle of mass `1 kg` moving in `x-y` plane obeys the law`U = 3x + 4y`, where `(x,y)` are the co-ordinates of the particle in metre. If the particle is at rest at `(6,4)` at time `t = 0` then :A. its acceleration is of magnitude `5m//s^(2)`B. Its velocity when it crosses the y axis is `10 m//s`C. it crosses the y - axis (x=0) at `y=-4`D. It moves in a straight line passing through the origin (0,0)
Answer» Correct Answer - A::B::C

Discussion

148.	A block of mass`2` kg is from to move along the x - axis it is at rest and from `1 = 0` onwards it is subjeted to a time - depended force `F(i) ` in the `x` diretion . The force `F(1) ` varies with `1` as shown in the figure . The kinetic of the block after `4.5` second is A. `4.50 J`B. `7.50 J`C. `5.06 J`D. `14.06 J`
Answer» Correct Answer - C If `u` is the final velocity, then change in momemtume `=` impulse `=` area enclosed by graph with time axis `2(upsilon-0)=(4xx3)/(2)-(1)/(2)xx(3)/(2)xx2=4.5` `:. upsilon=2.25 m//s` `KE=(1)/(2)m upsilon^(2)=(1)/(2)xx2xx(2.25)^(2)=5.06J`

Discussion

149.	A block of mass 30.0 kg is being brought down by a chain. If the block acquires as speed of `40.0 `cm/s` in dropping down 2.00 m ,find the work done by the chain during the process.
Answer» `W("all")=DeltaK =K_(f)-K_(i) =K_(f)` `:. W_(mg) + W_("chain")=K_(f)` or `W_("chain") =K_(f)-W_(mg) =1/2mv_(f)^(2)-mgh` `=1/2 xx 30 xx (0.4)^(2)-30 xx 10 xx 2` `=-597.6J`.

Discussion

150.	The potential energy funtions for the force between two along in a distance molecule is approximatily given by `U(x) = (a)/(x^(12)) - b)/(x^(6)) ` where `a` and `b` are constant and `x` is the distance between the aloms , if the discision energy of the molecale is `D = [U(x = oo) - U` atequlibrium ] , D isA. `(b^(2))/(12a)`B. `(b^(2))/(4a)`C. `(b^(2))/(6a)`D. `(b^(2))/(2a)`
Answer» Correct Answer - B `U_((x))-(a)/(x^(12))-(b)/(x^(6))` `F=-(dU(x))/(dx)=-(12a)/(x^(13))+(6b)/(x^(7))` At equilibrium distance between the molecules, `F=0` `:. -(12a)/(x^(13))+(6b)/(x^(7))=0` or `a=((2a)/(b))^(1//6)` From `(i)` `U_((oo))=(a)/(oo^(12))-(b)/(oo^(6))=0` Dissociation energy of molecule, `D=U_((x=oo))-U_("at equilibrium")` `=0-((a)/(x^(12))-(b)/(x^(6)))` where`x=((2a)/(b))^(1//6)` `D=(-a)/(x^(12))+(b)/(x^(6))=-(a)/((2a//b)^(12//6))+(b)/((2a//b)^(6//6))` `=-(b^(2))/(4a)+(b^(2))/(2a)=(b^(2))/(4a)`

Discussion

Explore topic-wise InterviewSolutions in .

An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring isA. `4 Mg//k`B. `2Mg//k`C. `Mg//k`D. `Mg//2k`

A block of mass `2kg` is propped from a heught of `40cm` on a spring where force constant is `1960Nm^(-1)` The maximum distance thought which the spring compressed byA. 5 cmB. 15 cmC. 20 cmD. 10 cm

A block of mass 0.1 kg attached to a spring of spring constant 400 N/m is putted rightward from `x_(0)=0` to `x_(1)=15` mm. Find the work done by spring force.

Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetricaly tostretch the spring by a length x over its natural length. The work doen by the spring oneach mass isA. `(1)/(2)kx^2`B. `-(1)/(2)kx^2`C. `(1)/(4)kx^2`D. `-(1)/(4)kx^2`

One end of a spring is rigidly fixed. A block attached to the free end of the spring is pulled through a distance `x_(0)`. On releasing the block, its amplitude of motion cannot exceed `+-x_(0)`. Why?

Two equal masses are attached to the two ends of a spring of spring constant `k`. The masses are pulled a part symmetrically to stretch the spring by a length `x` over its natural length. The work done by the spring on each mass is.A. `1/2 kx^(2)`B. `-1/2 kx^(2)`C. `1/4 kx^(2)`D. `-1/4 kx^(2)`

A ball of mass m is thrown in air with speed `v_(1)` from a height `h_(1)` and it is at a height `h_(2)(gth_(1))` when its speed becomes `v_(2)` . Find the work done on the ball the air resistance.

A man is riding on a cycle with velocity `7.2(km)/(hr)` up a hill having a slope 1 in 20. The total mass of the man and cycle is 100kg. The power of the man isA. 200 WB. 175 WC. 125 WD. 98W

A block of mass `m` at rest is acted upon by a force `F` for a time t. The kinetic energy of block after time t isA. `(Ft^2)/(m)`B. `(F^2t^2)/(2m)`C. `(Ft^2)/(2m)`D. `(F^2t^2)/(3m)`

How much work is done by a coolie walking on a horizontal platform with a load on his head ? Explain.

`300 J`of work is done in slide a `2kg` block up an inclined plane of height `10m`. `Taking g = 10 m//s^(2), work done against friction isA. `200 J`B. `100 J`C. zeroD. `1000 J`

300J of work is done in slinding a 2 kg block up an inclined plane of height 10m. Taking g =`10m//s^(2)` , work done against friction isA. 200JB. 100JC. zeroD. 1000J

A `1-KW` motor pumps out water from a well `10m` deep. Calculate the quantity of water pumped out per second.

(a) What is the power of an engine which can lift 600 kg of coal per minut from a mine 20 m deep ? (b) A 2kW motor pumps out water from a well `10m` deep. Calculated the quantity of water pumped out per second.

Calculate the power of a crane in watts, which lifts a mass of `100 kg` to a height of `10 m` in `20 s`.

A crane lifts a mass of 100 kg to a height of 10 m in 20 s. The power of the crane is (Take `g = 10 m s^(-2)`)A. 100 WB. 200 WC. 250 WD. 500 W

A crane lifts a mass of `100 kg` to a height of `10m` in `20s`. The power of the crane is `(Take g = 10 ms^(-2))`

Initially the system shown in figure is in equilibrium. At the moment, the string is cut the downward of `a(1)` and `a_(2)` are. .A. (a) zero and zeroB. (b) 2g and zeroC. (c) g and zeroD. (d) None of the above

A spring of force constant `k` is cut into there equal part what is force constant of each part ?

A spring of force constant k is cut in two parts at its one-third lingth. When both the parts are stretched by same amount. The word done in the two parts will be .A. equal in bothB. greater for the longer partC. greater for the shorter for the shorter partD. data insufficient.

A particle moving in a straight line is acted upon by a force which works at a constant rate and chages ist melocity from (u and v ) over a distance x. Prove that the taken in it is `3/2 (u+v)x/(u^(2)+v^(2)+uv)` .

The system is released from rest with the spring initally stretched `75 mm`. Calculate the velocity `v` of the block afrer it has dropped `12 mm`. The spring has a stiffness of `1050 N//m`. Neglect the mass of the small pulley.

With what minimum speed `v` must a small ball should be pushed inside a smooth vertical tube from a height `h` so that it may reach the top of the tube? Radius of the tube is `R` . A. `sqrt(2g(h + 2R))`B. `(5)/(2)R`C. `sqrt(g(5R - 2h))`D. `sqrt(2g(2R - h))`

A particle of mass `m` is released from a height `H` on a smooth curved surface which ends into a vertical loop of radius `R`, as shown. Choose the correct alernative(s) if `H=2R`. A. `H =(3R)/(2)`B. `H = 5R`C. `H =(5R)/(2)`D. None of these

The potential energy of a system increased if work is doneA. by the system against a conservative forceB. by the system against a non conservative forceC. upon the system by a conservative forceD. upon the system by a non conservative force

An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of `2 ms^(1)`. The mass per unit length of water in the pipe is `100 kgm^(-1)`. What is the power of the engine?A. 400 WB. 200WC. 100WD. 800W

force `F` on a partical moving in a straight line veries with distance `d` as shown in the figure. The work done on the partical during its displacement of `12 m` is A. 21JB. 26JC. 13JD. 18J

The kinetic energy of a body moving along a straight line varies with time as shown in figure. The force acting on the body: A. zeroB. constantC. directly proportional to velocityD. inversely proportional to velocity

The potential energy of a partical veries with distance `x` as shown in the graph. The force acting on the partical is zero atA. CB. BC. B and CD. A and D

force `F` on a partical moving in a straight line veries with distance `d` as shown in the figure. The work done on the partical during its displacement of `12 m` is A. `21 J`B. `26 J`C. `13 J`D. `18 J`

Velocity-time graph of a particle of mass (2 kg) moving in a straight line is as shown in Fig. 9.20. Find the word done by all the forces acting on the particle. .A. `400 J`B. `-400 J`C. `-200 J`D. `200 J`

The displacement of a body of mass `2 kg` veries with time `t` as `S = t^(2) + 2t`, where `S` is in seconds. The work done by all the forces acting on the body during the time interval `t = 2s` to `t = 4s` isA. `36 J`B. `64 J`C. `100 J`D. `120 J`

Given that the position of the body in `m` is a function of time as follows : `x = 2t^(4) + 5t + 4` The mass of the body is `2 kg` . What is the increase in its kinetic energy onesecond after the stert of motion?A. `168 J`B. `169 J`C. `32 J`D. `144 J`

A block of mass 30.0 kg is being brought down by a chain. If the block acquires as speed of `40.0 `cm/s` in dropping down 2.00 m ,find the work done by the chain during the process.