1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 5 InterviewSolution

201.	A baloon filled with helium rises against gravity increasing its potential energy. The speed of the baloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy ? You can neglect viscous drag of air and assume that density of air is constant.
Answer» LET m=Mass of ballon V=Volume of balloon `rho_(He)`=Density of helium `rho_("air")`=Density of air Volume V of ballon displaces volume V of air. So, `V(rho_("air")-rho_(He)g=ma=m(dv)/(dt)="up thrust"` Integrating with respect to t. `V(rho_("air")-rho_(He)gt=mv` `(1)/(2)mv^(2)=(1)/(2)m(V^(2))/(m^(2))(rho_("air")-rho_(He))^(2)g^(2)t^(2)` `(1)/(2m)V^(2)(rho_("air")-rho_(He))g^(2)t^(2)` If the ballon rises to a height h, from s=`ut+(1)/(2)at^(2)` `"We get" h=(1)/(2)at^(2)=(1)/(2)(V(rho_("air")-rho_("He")))/(m)"gt"^(2)` =`V(rho_("air")-rho_("He")gh` Rearranging the terms `(1)/(2)mv^(2)+Vrho_("He") gh=V_("pair")hg` `KE_("ballon")+PE_("ballon")="Change in PE of air"` So, as the ballon goes up , an equal volume of air comes down, increase in PE and KE of the ballon is the ballon is at the cost of PE of air [Which comes down]

Discussion

202.	Negative gradient of potential energy gives (a) conservative force (b) non conservative force (c) kinetic energy(d) frictional force
Answer» (a) conservative force

Discussion

203.	For perfectly inelastic collision, coefficient of restitution is (a) 0 (b) 1 (c) 0 < e < 1 (d) ∞
Answer» Correct answer is (a) 0

Discussion

204.	A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air is constant.
Answer» As dragging viscous force of air on balloon is neglected so there is Net Buoyant Force = Vρg Let m, V, ρ_HE, denote respectively the mass, volume and density of helium balloon and ρ_air density of air Volume V of balloon displaces volume V of air. So, V(ρ_air − ρ_HE)g = ma, or, V(ρ_air − ρ_HE)g = \(m\frac{dv}{dt}........(1)\) Integrating equation (1) with respect to t, we have V(ρ_air − ρ_HE)gt = mv \(\frac{1}{2}mv^2=\frac{1}{2}m\frac{v^2}{m^2}\)(ρ_air − ρ_HE)²g²t² \(=\frac{1}{2m}V^2\)(ρ_air − ρ_HE)²g²t² ......(2) If the balloon rises to a height h from h = ut + \(\frac{1}{2}\)at² We get, \(h=\frac{1}{2}at^2=\frac{1}{2}\frac{V(ρair − ρHE)}{m}gt^2\) (∵ u = 0).........(3) From Eqs. (3) and (2) \(\frac{1}{2}mv^2=[V(ρ_{air} − ρ_{HE})g][\frac{1}{2m}V(ρ_{air} − ρ_{HE})gt^2]\) = V(ρ_air − ρ_HE)gh Rearranging the terms. ⇒ \(\frac{1}{2}mv^2+Vρ_{HE}gh=Vρ_{air}gh\) Or KE_balloon + PE_balloon = changing in PE of air. So, as the balloon goes up, and equal volume of air comes down, increase in PE and KE of the balloon is at cost of PE of air [which comes down].

204.

A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air is constant.

Answer»

As dragging viscous force of air on balloon is neglected so there is Net Buoyant Force = Vρg

Let m, V, ρ_HE, denote respectively the mass, volume and density of helium balloon and ρ_air density of air

Volume V of balloon displaces volume V of air.

So,

V(ρ_air − ρ_HE)g = ma, or,

V(ρ_air − ρ_HE)g = \(m\frac{dv}{dt}........(1)\)

Integrating equation (1) with respect to t, we have

V(ρ_air − ρ_HE)gt = mv

\(\frac{1}{2}mv^2=\frac{1}{2}m\frac{v^2}{m^2}\)(ρ_air − ρ_HE)²g²t²

\(=\frac{1}{2m}V^2\)(ρ_air − ρ_HE)²g²t² ......(2)

If the balloon rises to a height h from h = ut + \(\frac{1}{2}\)at²

We get,

\(h=\frac{1}{2}at^2=\frac{1}{2}\frac{V(ρair − ρHE)}{m}gt^2\)

(∵ u = 0).........(3)

From Eqs. (3) and (2)

\(\frac{1}{2}mv^2=[V(ρ_{air} − ρ_{HE})g][\frac{1}{2m}V(ρ_{air} − ρ_{HE})gt^2]\)

= V(ρ_air − ρ_HE)gh

Rearranging the terms.

⇒ \(\frac{1}{2}mv^2+Vρ_{HE}gh=Vρ_{air}gh\)

Or KE_balloon + PE_balloon = changing in PE of air.

So, as the balloon goes up, and equal volume of air comes down, increase in PE and KE of the balloon is at cost of PE of air [which comes down].

Discussion

205.	If the velocity of separation is equal to the velocity of approach, then the collision is (a) conservative force (b) non conservative force (c) gravitational force (d) electrostatic force
Answer» (a) conservative force

Discussion

206.	Inelastic collision is due to(a) conservative force (b) non conservative force (c) gravitational force (d) electrostatic force
Answer» (b) non conservative force

Discussion

207.	The potential energy of a system increases, if work is done (a) by the system against a conservative force (b) by the system against a non-conservative force (c) upon the system by a conservative force (d) upon the system by a non-conservative force
Answer» (a) by the system against a conservative force

Discussion

208.	In which direction is the force of gravity on the body?
Answer» Direction of the gravitational force will be in the downward direction

Discussion

209.	If the work done is completely recoverable, then the force is (a) conservative (b) non-conservative (c) both (a) and (b) (d) frictional in nature
Answer» (b) non-conservative

Discussion

210.	Non conservative force is (a) frictional force (b) viscous force (c) air resistance (d) all the above
Answer» (d) all the above

Discussion

211.	The displacement is opposite to the frictional force, is the work done by the frictional force positive or negative
Answer» If the displacement is in opposite direction work done by frictional force is negative.

Discussion

212.	Write whether this work is negative or positive.
Answer» Work is positive

Discussion

213.	A particle moves with a velocity `6hat(i)-4hat(j)+3hat(k)m//s` under the influence of a constant force `vec(F)= 20hat(i)+15hat(j)-5hat(k)N`. The instantaneous power applied to the particle isA. `35(J)//(s)`B. `45(J)//(s)`C. `25(J)//(s)`D. `195(J)//(s)`
Answer» Correct Answer - B `P=vecF.vecv=(20hati+15hatj-15hatk).(6hati-4hatj+3hatk)` `=120-60-15=45W`

Discussion

214.	An ideal spring with spring - constant `K` is bung from the colling and `a` block of mass `M` is attached to its lower end the mass is released with the spring initally unstetched . Then the maximum exlemsion in the spring isA. `(4Mg)/(K) `B. `(2Mg)/(K) `C. `(Mg)/(K) `D. `(Mg)/(4K) `
Answer» Correct Answer - B (b) The shown situation can also be looked upon as the decreses is the gaviational potental energy of spring mass system is equal to the gain in spring elastic poential energy `Mgx = (1)/(2) kr^(2) , x = (2Mg)/(k)`

Discussion

215.	The kinetic energy acquired by a mass `m` travelling a certain distance d, starting from rest, under the action of a force F such that the force F is directly proportional to t isA. directly proportional to `t^(2)`B. independent of tC. directly proportional to `t^(4)`D. directly proportional to t
Answer» Correct Answer - C

Discussion

216.	The `K.E.` acquired by a mass `m` in travelling a certain distance `d`, starting from rest, under the action of a constant force is directly propotional toA. directly proportional to mB. directly proportional to `sqrtm`C. inverserly proportional to `sqrtm`D. independent of m
Answer» Correct Answer - D (d) `K_(2)-K_(1) = W =Fs` `K_(2)-0=Fs` K is independent of m. `" "`(Force is constant)

Discussion

217.	For a particle moving on a straight lint the variation of acceleration with time is given by the graph as shown. Initially the particle was at rest. Then the corresponding kinetic energy of the particle versus time graph will be .A. B. C. D.
Answer» Correct Answer - A

Discussion

218.	A wind - powered generator convets and energy into electrical energy . Assume that the generator convents a fixed fraction of the wind energy intercepited by to blades into electrical energy for wind speed `V` , the electrical power output will be propertional toA. `v`B. `v^(2)`C. `v^(3)`D. `v^(4)`
Answer» Correct Answer - C

Discussion

219.	A particle, which is constrained to move along the x-axis, is subjected to a force from the origin as `F(x) = -kx + ax^(3)`.Here `k` and a are positive constants. For `x=0`, the functional form of the potential energy `U(x)` of particle is.A. B. C. D.
Answer» Correct Answer - D

Discussion

220.	A particle, which is constrained to move along the x-axis, is subjected to a force from the origin as `F(x) = -kx + ax^(3)`.Here `k` and a are positive constants. For `x=0`, the functional form of the potential energy `U(x)` of particle is.A. (a) B. (b) C. (c) D. (d)
Answer» Correct Answer - D `dU_((x))=-Fdx` `:. U_x=underset0oversetx intFdx=(kx^2)/(2)-(ax^4)/(4)` `U=0` at `x=0` and at `x=sqrt((2k)/(a))` `U(x)` is negative for `xgtsqrt((2k)/(a))` `F=0` at `x=0`. It means slope of `U-x` graph is zero at `x=0`.

Discussion

221.	A body constrained to move along the z-axis of a co-ordinate system, is subjected to a constant force `vec(F)` given by `vec(F)=-hat(i)+2hat(j)+3hat(k)` Newton where `hat(i),hat(j)`and `hat(k)` represent unit vectors along x-,y-,and z-axes of the system, respectively. Calculate the work done by this force in displacing the body through a distance of `4m` along the z-axis.
Answer» Since the body is displaced `4m` along the z-axis only, `S=0hati+0hatj+4hatk` Also, `F=-i+2hatj+3hatk` Work done, `W=vecF.vecS=(-hati+2hatj+3hatk)*(0hati+hatj+4hatk)` `=12(hatk.hatk)J=12J`

Discussion

222.	A particle in a certain conservative force field has a potential energy given by `V=(20xy)/z`. The force exerted on it isA. `((20y)/z)hati+((20x)/z)hatj+((20xy)/z^2)hatk`B. `-((20y)/z)hati-((20x)/z)hatj+((20xy)/z^2)hatk`C. `-((20y)/z)hati-((20x)/z)hatj-((20xy)/z^2)hatk`D. `((20y)/z)hati+((20x)/z)hatj-((20xy)/z^2)hatk`
Answer» Correct Answer - B Given : `V=(20xy)/z` For a conservative field `vecF= -vecgradV, "where" vecgrad=hati del/(delx)+hatjdel/(dely)+hatkdel/(delk)` `therefore vecF= - [hati(delV)/(delx)+hatj(delV)/(dely)+hatk(delV)/(delz)]` `=-[hati del/(delx)((20xy)/z)+hatjdel/(dely)((20xy)/z)+hatkdel/(delz)((20xy)/z)]` `=-((20y)/z)hati-((20x)/z)hatj+((20xy)/z^2)hatk`

Discussion

223.	Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C and D is which the relation between potential energy U, kinetic energy (K) and total energy E is as given below RegionA:`UgtE` Region B:`UltE` Region C:`KltE` Region D:`UgtE` State with reason in each case whether a particle can be found in the given region or not.
Answer» We know that Total ME=KE+PE `Rightarrow E_(0)=KE+V(x)` ltbRgt `Rightarrow KE=E_(0)-V(x)` at `A_(1) x=0,V(x)=E_(0)` `Rightarrow KE=E_(0)-E_(0)=0` at `B_(1) V(x) gt E_(0)` `Rightarrow KE gt0` at C and `D_(1) V(x)=0` This is possible because total energy can be greater than PE(V). For region C Given, `K gt E Rightarrow K -E gt 0` From Eq. (i) PE=V=E-Klt0 Which is possible, because PE can be negative For region D Given,` V gt K` This is possible because for a system PE(V) may be greater than KE(K)

Discussion

224.	Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C and D is which the relation between potential energy U, kinetic energy (K) and total energy E is as given below RegionA:`UgtE` Region B:`UltE` Region C:`KltE` Region D:`UgtE` State with reason in each case whether a particle can be found in the given region or not.A. Region AB. Region BC. Region CD. Region D
Answer» Correct Answer - A Total energy of the particle = Kinetic energy + Potential energy E = K+ V or K=E-V Since kinetic energy can never be negative. The particle cannot be found in the region where its kinetic energy would become negative In region A, V gt E `therefore` K becomes negative in this region. Hence, the particle cannot be found in region A.

Discussion

225.	A raindrop of mass `1g` falling from a height of `1km` hits is the ground with a speed of `50 ms^(-1)`. Which of the following statements is correct? `("Taking" g= 10ms^(-2))`.
Answer» Given, mass of the rain drop(m)=100g =`1xx10^(-3)kg` Height of falling(h)=`1km=10^(3)m=g=10m//s^(2)` Speed of the rain drop(v)=50m/s (a) Loss of PE of the drop=mgh =`1xx10^(-3)xx10xx10^(3)=10J` (b) Gain in KE of the drop=`(1)/(2)mv^(2)` `(1)/(2)xx1xx10^(-3)xx(50)^(2)` `=(1)/(2)=xx10^(-3)xx2500=1.250J` (c ) No, gain in KE is not equal to the loss in its PE, because a part of PE is utilised in doing work against the viscous drag of air.

Discussion

226.	A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 ms-1. Calculate(a) The loss of P.E. of the drop.(b) The gains in K.E. of the drop.(c) Is the gain in K.E. equal to loss of P.E.? If not why.Take g = 10 ms-2
Answer» Given mass of rain drop, (m) = 0.001 kg = 1.0 × 10^-3 kg Height, h = 1 km = 1000 m Speed, v = 50 m/s, u = 0. (a) Loss of PE = mgh = 1 × 10⁻³ × 10 × 10³ = 10 J (b) Gain in KE = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × 10⁻³ × 2500 = 1.25 J (c) No, Because a part of PE is used up in doing work against the viscous drag of air.

226.

A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 ms-1. Calculate(a) The loss of P.E. of the drop.(b) The gains in K.E. of the drop.(c) Is the gain in K.E. equal to loss of P.E.? If not why.Take g = 10 ms-2

Answer»

Given mass of rain drop, (m) = 0.001 kg

= 1.0 × 10^-3 kg

Height, h = 1 km = 1000 m

Speed, v = 50 m/s, u = 0.

(a) Loss of PE = mgh = 1 × 10⁻³ × 10 × 10³ = 10 J

(b) Gain in KE = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × 10⁻³ × 2500 = 1.25 J

(c) No, Because a part of PE is used up in doing work against the viscous drag of air.

Discussion

227.	A raindrop of mass `1g` falling from a height of `1km` hits is the ground with a speed of `50 ms^(-1)`. Which of the following statements is correct? `("Taking" g= 10ms^(-2))`.A. The loss of potential energy of the drop is 10 JB. The gain in kinetic energy of the drop is 1.25 JC. The gain in kinetic energy of the drop is not equal to the loss of potential energy of the drop.D. All of these
Answer» Correct Answer - D Here, m=1 `g= 10^(-3)` kg, h = 1 km = 1000 m `g = 10 m s^(-2), v = 50 m s^(-1)` Loss of potential energy of the drop `=mgh=(10^(-3) kg)(10 m s^(-2)) (1000 m) = 10 J`...(i). Hence, option (a) is correct. Gain in kinetic energy of the drop `=1/2mv^2=1/2xx(10^(-3) kg)(50 m s^(-1))^2=1.25 J` Hence, option (b) is correct. From (i) and (ii), we get Loss of potential energy of the drop `ne` Gain in kinetic energy of the drop. Hence, option (c) is also correct.

Discussion

228.	Suppose the average mass of raindrops is `3.0xx10^(-5)kg` and their average terminal velocity `9ms^(-1)`. Calculate the energy transferred by rain to each square metre of the surface at the place which receives 100 cm of rain in a year.
Answer» Given average mass of rain drop=`(m)=3.0xx10^(-3)xxkg` Average terminal velocity`=(V)=9m//s.` Height(h)=100cm-1m Density of water `(rho)=10^(3)kg//m^(3)` Area of the surface(A)=`1m^(2)` Volume of water due to rain(V)=`"Area"xx"height"` `-Axxh` `=1xx1=1m^(3) Mass of the water due to rain (M)=`"Volume"xx"densitt"` `=Vxxrho` `=1xx10^(3)` `=10^(3)kg` Energy transferred to the surface=`(1)/(2)mv^(2)` `(1)/(2)xx10^(3)xx(9)^(2)` `=40.5xx10^(3)J=4.05xx10^(4)J`

Discussion

229.	Suppose the average mass of raindrops is 3.0 × 10-5 kg and their average terminal velocity 9 m s-1. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.
Answer» Give : m = 3.0 × 10^-5 kg, ρ = 1.0 × 10³ kg/m³ v = 9 m/s A = 1m² h = 100 cm ⇒ V = 1m³ M = ρV = 10³ kg, Energy transferred by rain, E = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × 10³ × (9)³ = 4.05 × 10⁴ J.

229.

Suppose the average mass of raindrops is 3.0 × 10-5 kg and their average terminal velocity 9 m s-1. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.

Answer»

Give : m = 3.0 × 10^-5 kg,

ρ = 1.0 × 10³ kg/m³

v = 9 m/s

A = 1m²

h = 100 cm ⇒ V = 1m³

M = ρV = 10³ kg,

Energy transferred by rain,

E = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) × 10³ × (9)³ = 4.05 × 10⁴ J.

Discussion

230.	Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position, the two bobs are in constact, . 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 pendulum with time,for `0letle2T` , where `T` is the period of each pendulum.
Answer» As two pendulum have identical bobs and lengths, they will execute simple harmonic motions of same time period, say `T`. At `t=0`, suppose bob `A` is displaced by `10^(@)` to the left. It is given potential energy `E_(1)=E` . Energy of `b,e_(2)=0`. When `A` is released, it strikes `B` at `t=T//4` . In the head on elastic collision between `A` and `B`, `A` comes to rest and `B` gets velcoity of `A`. Therefore, `E_(1)=0` and `E_(2)=E`. At `t=2T//4`, `B` reahes its extreme right position when `K.E.` of `B` is converted into `P.E.=E_(2)=E`. Energy of `A,E_(1)=0.`At `t=3T//4,` `B` reaches its mean position, when its P.E. is converted into `KE.=E_(2)=E`. It collides elastically with `A` and transfers whole of its energy to `A` . Thus, `E_(2)=0` and `E_(1)=E`. The entire process is repeated. The value of energies of `A` and `B` at different time intervals are tabulated here. The plot of energy with time `0letle2T` is shown separately for `A` and `B` in figure. `\|{:(time(t),"Energy of" A,"Energy of" AB,),(,E_(1),E_(2),),(0,E,0,),(T//4,0,E,),(2T//4,0,E,),(3T//4,E,0,),(4T//4,E,0,),(5T//4,0,E,),(6T//4,0,E,),(7T//4,E,0,),(8T//4,E,0,):}\|`

Discussion

231.	In unloading grain from the hold of a ship, an elevator lifts the grian through a distance of 12 m. Grain is discharged at the top of the elevator at a rate of 2 kg each second and the discharge speed of each of grain particle is 3 `ms^(-1)`. Find the minimum horsepower of the motor that can elevate grain in this way. (Take g=10 `ms^(-2)`)
Answer» The work done by the motor each second ,i.e., Power`=mgh+(1)/(2)mv^(2)`, as t=1 s. Given, m=2kg, `v=3 ms^(-1)` and h=12 m. `:. "Power"=2xx(10)xx(12)+(1)/(2)(2)(3)^(2)=249 W=(249)/(746)=0.33 hp` The motor must have an output of at least 0.33 hp.

Discussion

232.	A uniform chain of mass m and length l is lying on a horizontal table with one-third of its length hanging over the edge of the table. If the chain is limiting equlibrium, what is the coefficient of friction for the contact between table and chain?A. `1/2`B. `1/3`C. `2/3`D. `3/2`
Answer» Correct Answer - A

Discussion

233.	A uniform chain of mass m and length l is lying on a horizontal table with one-third of its length hanging over the edge of the table. If a gentale pull is given to hanging end, it will start moving down. What is the change in gravitational potential energy of the chain, as it jist leaves the table?A. `(4mgl)/(9)`B. `-(4mgl)/(9)`C. `(17mgl)/(18)`D. `-(mgl)/(18)`
Answer» Correct Answer - B

Discussion

234.	If the net work done by external force on a partical is zero which is the following statement about the partical must be true?A. Its velocity is constantB. Its velocity is decreasedC. Its velocity is unchangedD. Its speed is unchanged
Answer» Correct Answer - D The work - energy theorm states that `W_("net") = Delta K = K_(f) - K_(i)` Thus `W_("net") = 0` then `K_(f) - K_(i) or (1)/(2) mv_(f)^(2) - (1)/(2) mv_(i)^(2)`, which leads to the conclusion that the speed is unchanged `(v_(f) - v_(i))`. The velocity of the partical involves both magnitude (speed) and direction. The work energy theorem shows that the magnitude or speed is unchanged when `W_("net") = 0` but makes no statement about the direction of the velocity.

Discussion

235.	Is the work required to be done by an external force on an object on a frictionless , horizontal surface to accelerate it from a speed `v` to a speed `2y`A. equal to the work required to accelerate the object from `t = 0` to `v`.B. twice the work required to accelerate the object from `v = 0` to `v`.C. three time the work required to accelerate the object from `v = 0` to `v`.D. four time the work required to accelerate the object from ` 0` to `v`. Or
Answer» Correct Answer - C The net work needed to accelerate the object from `v= 0` to `v` is `W_(1) = KE_(1 f) - KE_(1 i) = (1)/(2) mv^(2) - (1)/(2) m (0) ^(2) = (1)/(2) mv^(2)` The work required to accelerate the object from speed `v` to speed `2 v` is `W_(2) = KE_(2 f) - KE_(2 i) = (1)/(2) m(2 v)^(2) - (1)/(2) mv^(2)` ` = (1)/(2) m (4 v^(2) - v^(2)) = 3 ((1)/(2) mv^(2)) = 3 W_(1)`

Discussion

236.	Two identical cylindrical vessel with their bases at the same level each contain a liquid of density `rho`. The height of the liquid in one vessel is `h_(1)` and in the other is `h_(2)` the area of either base is A. What is the work done by gravity is equalising the levels when the two vessels are connected?A. `(h_(1)-h_(2))grho`B. `(h_(1)h_(2))gArho`C. `(1)/(2)(h_(1)-h_(2))^(2)gArho`D. `(1)/(4)(h_(1)-h_(2))^(2)gArho`
Answer» Correct Answer - D

Discussion

237.	A body moves a distance of 10 m along a straight line under the action of a force of 5 N. If the work done is 25 joules, the angle which the force makes with the direction of motion of the body isA) 0° B) 30° C) 60° D) 90°
Answer» C) 60° Explanations: W = F s cosθ ⇒ cosθ= W/F_s= 25/50=1/2 θ=60°

237.

A body moves a distance of 10 m along a straight line under the action of a force of 5 N. If the work done is 25 joules, the angle which the force makes with the direction of motion of the body isA) 0° B) 30° C) 60° D) 90°

Answer»

C) 60°

Explanations:

W = F s cosθ

⇒ cosθ= W/F_s= 25/50=1/2

θ=60°

Discussion

238.	A ball of mass m is dropped from a cliff of height H. The ratio of its kinetic energy to the potential energy when it is fallen through a height 3/4 H isA. `3:4`B. `4:3`C. `1:3`D. `3:1`
Answer» Correct Answer - C Total mechanical energy at height, H `E_H=mgH` Let `v_h` be velocity of the ball at height h `(=3/4H)` `therefore` Total mechanical energy at height h, `E_h=mgh+1/2mv_h^2` According to law of conservation of mechanical energy, `E_H=E_h, mgH = mgh+1/2mv_h^2` `v_h^2=2g(H-h)` Required ratio of kinetic energy to potential energy at height h is `K_h/V_h=(1/2mv^2h)/(mgh) =(1/2m2g(H-h))/(mgh)=(H/h-1)=1/3`

Discussion

239.	A man `M_(1)` of mass `80 kg` runs up a statircase in `15 s`. Another man `M_(2)` also of mass `80 kg` runs up the same staricase in `20 s`. The ratio of the power devlopment by then will be:A. 1B. `(4)/(3)`C. `(16)/(9)`D. none of these
Answer» Correct Answer - B Power `P=` Work time Work done by the will be same . Hence `(P_(1))/(P_(2)) = (t_(2))/(t_(1)) = (20)/(15) = (4)/(3)` .

Discussion

240.	A body of mass `m` is acceleratad uniformaly from rest to a speed `v` in a time `T` . The instanseous power delivered to the body as a function of time is given byA. `(1)/(2) m (v)/(t_(1)) t^(2)`B. `m (v)/(t_(1)) t^(2)`C. `(1)/(2) ((mv)/(t_(1)))^(2) t^(2)`D. `(1)/(2) m (v^(2))/(t_(1)^(2)) t^(2)`
Answer» Work done `= F xx s` `= ma xx (1)/(2) at^(2) [From x = ut + (1)/(2) at^(2)]` `:. W = (1)/(2) ma^(2) t^(2) = (1)/(2) m ((v)/(t_(1)))^(2) t^(2) [As a = (v)/(t_(1))]`

Discussion

241.	A body of mass `m` is acceleratad uniformaly from rest to a speed `v` in a time `T` . The instanseous power delivered to the body as a function of time is given byA. `(V)/(T) t`B. `(V^(2))/(T) t^(2)`C. `(V^(2))/(T^(2)) t`D. `(V^(2))/(T^(2)) t^(2)`
Answer» Correct Answer - C Power `= ("work done")/(time) = (V^(2) t^(2))/(T^(2) t)` Power prop `(V^(2) t)/(T^(2)`.

Discussion

242.	When slow neutrons are incident on a target containing `._(92)U^(235)` , a possitle fission reactionis `._(92)U^(235)+n rarr ._(56)Ba^(141)+_(36)Kr^(92)+3n+Q` Estimate the amount of energy released using the following data : `M[._(92)U^(235)]=235.04u,` `M[._(56)Ba^(141)]=140.91u,` `M[._(36)Kr^(92)]=91.926u,` `M_(n)=1.0087u.` Take `1u=1.661xx10^(-27)kg.` `1MeV=1.602xx10^(-13)J`
Answer» Mass defect, `Deltam=235.04+1.0087` `-140.91-91.926-3xx1.0087` `Deltam=0.1866u=0.1866xx1.661xx10^(-27)kg` Energy released, `E=(Deltam)c^(2)` `=0.1866xx1.661xx10^(-27)(3xx10^(8))^(2)J` `=(0.1866xx1.661xx9xx10^(-11))/(1.602xx10^(-13))MeV` `E=174.1MeV`

Discussion

243.	A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in figure. Correctly shows the displacement-time curve for its motion ?A. B. C. D.
Answer» Correct Answer - B For constant power, displacement `prop t^(3//2)`

Discussion

244.	A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in figure. Correctly shows the displacement-time curve for its motion ?A. B. C. D.
Answer» (b) Given power=constant We know that powre (P) `P=(dW)/(dt)=(F.ds)/(dt)=(F.ds)/(dt) (therefore "body is moving undirectly")` Hence, `F.ds=Fds cos0^(@)` `P=(Fds)/(dt)="constant"` Now, writing dimensions lt`[F][v]="constant"` `Rightarrow [MLT^(-2)][LT^(-1)]="constant"` `Rightarrow L^(2)T^(-3)="constant" (therefore "mass of constant")` `Rightarrow L alpha T^(3//2) Rightarrow "Displacement(d) alphat^(3//2)`

Discussion

245.	A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional tot1/2 t t3/2 t2
Answer» It is proportional ‘t^3/2’ We know that P = F × V [P] = [F] [V] [P] = [MLT^-2 ] [LT^-1 ] since ‘P’ & ‘M’ are constant. L²T^-3 = constants ⇒ \(\frac{L^2}{T^3}\)= constant. L²& T³ ⇒ L & T^3/2.

245.

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional tot1/2 t t3/2 t2

Answer»

It is proportional ‘t^3/2’

We know that P = F × V

[P] = [F] [V]

[P] = [MLT^-2 ] [LT^-1 ] since ‘P’ & ‘M’ are constant.

L²T^-3 = constants

⇒ \(\frac{L^2}{T^3}\)= constant.

L²& T³ ⇒ L & T^3/2.

Discussion

246.	A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in figure. Correctly shows the displacement-time curve for its motion ?A. B. C. D.
Answer» Correct Answer - B Here, `P=[ML^(2)T^(-3)]=`constant `(L^(2))/(T^(3))=`constant or `LpropT^(3//2)` or displacement`(d)propt^(3//2)`

Discussion

247.	A mass of `50 kg` is raised through a certain height by a machine whose efficiency is `90%` , the energy spend is `5000 J`. If the mass is now released, its `KE` on hitting the ground shall be:A. `5000 J`B. `4500 J`C. `4000 J`D. `5500 J`
Answer» Because the efficiency of machine is `90%`, hence potential energy gained by mass `= (90)/(100) xx energy = (90)/(100) xx 5000 J = 4500 J` When the mass is released now, gain in `KE` on hitting the ground = loss of potential energy =` 4500 J`

Discussion

248.	The potential energy function associated with the force `vecF=4xyhati+2x^2hatj` isA. (a) `U=-2x^2y`B. (b) `U=-2x^2y+const ant`C. (c) `U=2x^2y+const ant`D. (d) Not defined
Answer» Correct Answer - B `dU=-vecF.ds=-vecF.(dxhati+dyhatj)` Also by reverse method using `F_x=(delU)/(delX)` and `F_y=-(delU)/(delY)` only option(b) satisfies the criteria.

Discussion

249.	A body of mass m moving with a constant velocity v hits another body of the same mass moving with the same velocity v but in the opposite direction and sticks to it. The velocity of the compound body after collision isA. vB. 2vC. zeroD. v/2
Answer» Correct Answer - C

Discussion

250.	A ball hits a vertical wall horizontal at 10 m/s bounces back at 10 m/sA. thereis no acceleration because `10(m)/(s)-10(m)/(s)=0`B. There may be an acceleration because its initial direction is horizontalC. There is an acceleration because there is a momentum changeD. Even though there is no change in momentum there is a change in direction. Hence it has an acceleration
Answer» Correct Answer - C

Discussion

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