Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

MCQ Questions for Class 11 Medical Physics Work, Energy And Power Quiz 10

A particle of mass 1 kg is projected under gravity at an angle

${ 37 }^{ o }$ with horizontal with 50 m/s. Instantaneous power delivered by gravity at t=1s is

$\left( sin{ 37 }^{ o }=\frac { 3 }{ 5 } ,cos{ 37 }^{ o }=\frac { 4 }{ 5 } \right)$

0%
200 J/s
0%
-200 J/s
0%
Zero
0%
-400 J/s

A blocks is moved from rest through a distance of

$4m$ along a straight line path. The mass of the blocks is

$5kg$ and the force acting on it is

$20N$ . If the kinetic energy acquired by the block be

$40J$ , at what angle to the path the force is acting-

0%
${30}^{o}$
0%
${60}^{o}$
0%
${45}^{o}$
0%
none of the above

Mass of

$B$ is four times that of

$A . B$ moves with a velocity half that of

$A$ . Then,

$B$ has

0%
kinetic energy equal to that of $A$
0%
half the kinetic energy of $A$
0%
twice the kinetic energy of $A$
0%
kinetic energy one-fourth of $A$

Explanation

According to question given,

$M_B = 4 \times M_A$ and

$V_B = \dfrac{V_A}{2}$

$K.E_A = \dfrac{1}{2}M_AV_A^2$ .........(1)

$K.E_B = \dfrac{1}{2}M_BV_B^2$ =

$\dfrac{1}{2}\ (4 \times M_A) \left(\dfrac{V_A}{2}\right)^2=\dfrac{1}{2}M_AV_A^2$ ........(2)

From above equation

$K.E_A=K.E_B$

The work done in placing a charge of 8

$\times$

${ 10 }^{ -18 }$ coulomb on a condenser of capacity 100 micro-farad is

0%
16 $\times { 10 }^{ -32 }$ joule
0%
3.2 $\times { 10 }^{ -26 }$ joule
0%
4 $\times { 10 }^{ -10 }$ joule
0%
32 $\times { 10 }^{ -32 }$ joule

A body of mass 1 kg begins to move under the action of a time dependent force

$\overrightarrow { F } =(2t\hat { i } +3t^{ 2 }\hat { j } )$ N, where

$\hat { i }$ and

$\hat { j }$ are unit vectors along x and y axes. What power will be developed by the force at the time t?

0%
$(2t^2+3t^2)W$
0%
$(2t^4+4t^4)W$
0%
$(2t^3+3t^4)W$
0%
$(2t^3+3t^5)W$

Explanation

As given

$\overrightarrow{F}=(2t\hat{i}+3t^2\hat{j})N$

Where

$\hat{i}$ and

$\hat{j}$ are unit vectors along x and y axis

$F=ma$

$\Rightarrow a=\cfrac{F}{m}=\cfrac{(2t\hat{i}+3t^2\hat{j})}{2}=[m=1kg]\\ \Rightarrow a=(2t\hat{i}+3t^2\hat{j})m/s^2$

Acceleration

$a=\cfrac{dv}{dt}$

$\Rightarrow dv=a.dt\rightarrow(i)\\ \int{dv}=\int{a.dt}=\int{(2t\hat{i}+3t^2\hat{j})}dt\\v=t^2\hat{i}+t^3\hat{j}\\P=F.v=(2t\hat{i}+3t^2\hat{j})(t^2\hat{i}+t^3\hat{j})\\ \quad=2t.t^+3t^2t^3\\P=(2t^3+3t^5)W$

A wooden plank of length

$L$ , reacts on a frictionless floor. A boy of mass

$M$ now runs over the plank starting from its one end. If mass of wooden plank is

$M/5$ , the distance covered by the boy relative to the ground will be:

0%
$L/6$
0%
$5L/6$
0%
$L/5$
0%
$4L/5$

A block of mass 10 kg is moving up on an inclined plane with constant acceleration 2 m/sec by applying a constant force F as shown in the fig. The work done on the block by applied force F in 2 s is

0%
480
0%
240
0%
960
0%
none

$\overrightarrow { F } =x\hat i+y\hat j$ , is this force:

0%
a non conservation
0%
a conservation force
0%
depends upon X
0%
depends upon $y \quad \& \quad x$

A force of

$1200 \mathrm { N }$ acting on a stone by means of a rope slides the stone through a distance{ of

$10 \mathrm { m }$ in a direction inclined at

$60 ^ { \circ }$ force. The work done by the force is

0%
$6000 \sqrt { 3 } J$
0%
6000 J
0%
12000 J
0%
8000 J

Explanation

Hence, the option

$B$ is the correct answer.
1380140_1320760_ans_c629284accb34c459d054f61bd8c4a86.png

$\begin{array} { l } { \text { A man weighing } 80 \mathrm { kg } \text { climbs a staircase } } \\ { \text { carrying a } 20 \mathrm { kg } \text { load. The staircase has } 40 } \\ { \text { steps, each of } 25 \mathrm { cm } \text { height. If he takes } 20 } \\ { \text { seconds to climb, the work done is } } \end{array}$

0%
$9800 \mathrm { J }$
0%
$245 \mathrm { J }$
0%
$98 \times 10 ^ { 4 } J$
0%
7840 J

Explanation

As the man climb up the stair he does the work against the gravity

$,$ the total gravitational force act on it in downward direction

$=(80+20)g$

here

$g$ is the acceleration due to gravity

$,$

height raised by the man

$25cm=0.25m$

work done by the man

$\left( {80 + 20} \right) \times g \times 0.25 = 245J$

Hence,

option

$(B)$ is correct answer.

A variable force given by

$F= ( 3x^2 \hat i + 4 \hat j)$ acts on a particle. the force is in newton and x is in meter. what is the change in KE of particle as it moves from point (2, 3) to (3, 0)

0%
-7J
0%
zero
0%
7J
0%
19J

A particle moves along positive X-axis from x=

$0$ to x=

$5$ under influence of force is given by

$F={ x }^{ 2 }-2x-8.$ The work done by the force is

0%
-69 units
0%
-23.33 units
0%
-5 units
0%
zero

The displacement (x) and time (t) for a particle are related as

$t=\sqrt { x } +3$ . What is the work done in the first six second of its motion?

0%
3 J
0%
4 J
0%
Zero
0%
5 J

Explanation

Given,

$t=\sqrt x +3$

Then,

$\sqrt x=t-3 \implies x=t^2+9-6t$

Also,

$v=\dfrac{dx}{dt}= 2t-6$

$t=0, v=-6\,ms^{-1}$

$t=6\,sec, v=6\, ms^{-1}$

Work done = Change in K.E

$=K.E_f-K.E_i$

$=\dfrac 12 m(6)^2-\dfrac 12 m(6)^2=0$

Five identical balls each of mass m and radius r are strung like beads at random and at rest along a smooth, rigid horizontal thin rod of length L, mounted between immovable supports. Assume

$10r<L$ and that the collision between balls or between balls and supports are elastic. If one ball is struck horizontally so as to acquire a speed v, the average force felt by the support is

0%
$\dfrac { 5m }{ L-5r }$
0%
$\dfrac { m{ v }^{ 2 } }{ L-10r }$
0%
$\dfrac { 5m{ v }^{ 2 } }{ L-10r }$
0%
$\dfrac { m{ v }^{ 2 } }{ L-5r }$

Two bodies of masses

$4kg$ and

$16kg$ are at rest. The ratio of times for which the same force must act on them to produce the same kinetic energy in both of them is

0%
$1:4$
0%
$2:1$
0%
$1:2$
0%
$4:1$

Explanation

kinetic energy is given by

$:$

so taking velocity for mass

$1kg$ as

$v_1,$ we get

and taking velocity for mass

$4kg$ as

$v1,$ we have

given that

$e_1 = e_2,$

equating the above terms and simplifying, we get

$,$

now , momentum is given by

$,$

let us assume

$, p_1$ as momentum for mass

$1kg$ and

$p_2$ for mass

$4kg$

$,$

$p_1 = 1 × v_1$

and

$p_2 = 4×v_2$

then

$,$

$\dfrac{{{p_1}}}{{{p_2}}} = \dfrac{1}{4} = 1:4$

Hence,

option

$(A)$ is correct answer.

does not vary from point to point in space Which pair of the following forces will never give resultant force of 2 N?

0%
1 N and 3 N
0%
2 N and 2 N
0%
1 N and 1 N
0%
1 N and 4 N

Explanation

$1N$ and

$4N$ will never give

$2N$

mind it

$,$ if the side lengths of a triangle is

$a,b,c$

$a+b>=c$

$b+c>=a$

$c+a>=b$

Hence,

option

$(D)$ is correct answer.

A hollow smooth uniform sphere

$A$ of mass

$m$ rolls without sliding on a smooth horizontal surface. It collides elastically and head-on with another stationary smooth hollow sphere

$B$ of the same mass mm and same radius. The ratio of the kinetic energy of

$B$ to that of

$A$ just after the collision is

0%
$1 : 1$
0%
$2 : 3$
0%
$3 : 2$
0%
$\rm{None}$

If body of mass 2 kg is moving along X axis under an external force, whose displacement is given by

$x(t) = 3t^2 - 2t + 5 m$ . Find the work done by force from t = 0 to t = 5 sec.

0%
$78\quad J$
0%
$780\quad J$
0%
$840\quad J$
0%
$200\quad J$

The work done by a force $F=(-6x^3\hat i)$ in displacing a particle from $x = 4m$ to $x = -2m$ is:

0%
$- 240$ J
0%
$360$ J
0%
$420$ J
0%
will depend upon the path

A force

$\vec { F } = 2 x \hat { i } + 3 y ^ { 2 } \hat { j } \mathrm { N }$ where x and y are in meter, acts upon a particle. Find work done (in J) by this force in displacing the particle from (1 m, 2 m, 5 m) to (2 m, 3 m, 2 m).

0%
24
0%
16
0%
28
0%
22

A particle moves in a potential region given by

$U = 8{x^2} - 4x + 400\,J.$ Its state of equilibrium will be

$-$

0%
8 m
0%
10 m
0%
12 m
0%
0.25 m

Explanation

$U = 8x^2 - 4x + 400$
Differentiate with

$dx$ we will get

$F$

$F = 16x - 4$
At equilibrium

$F = 0$

$0 = 16x - 4$

$x = 1/4 = 0.25 m$
It’s state of equilibrium will be at

$x = 0.25 m$

Hence,

option

$(D)$ is correct answer.

Work done by air when it expands from 50 litres to 150 litres at a constant pressure of 2 atmospheres is:

0%
$2\times { 10 }^{ 4 }J$
0%
$2\times { 10 }0J$
0%
$2\times { 10 }^{ 5 }\times 100J$
0%
$2\times { 10 }.5\times 100J$

Two bodies traveling with velocity

$20ms^{-1}$ and

$40ms^{-1}$ have same K.E. If the mass of the body having velocity

$40ms^{-1}$ is 8 kg the mass of other body is

0%
8 Kg
0%
16 kg
0%
24 kg
0%
32 kg

Explanation

Given

$V_1 = 20 m/s$ and

$V_2 = 40 m/s$

$M_2 = 8\ kg$

$K.E_1 = K.E_2$

$\dfrac{1}{2}M_1V_1^2 = \dfrac{1}{2}M_2V_2^2$

$\implies M_1 = M_2 \times \dfrac{V_2^2}{V_1^2} = 8\times \dfrac{40^2}{20^2} = 32\ kg$

A body of mass 1 kg falls from rest through the air, a distance of 200 m and acquires a speed 50 m/s. Work done against air friction is

0%
$750 J$
0%
$1250 J$
0%
$1960 J$
0%
$3210 J$

A block of mass m is oscillating on smooth between two light springs of spring constant K separated by a distance I colliding elastically with the spring.If the velocity of the blocks is increased by an external impulse when it is not touching either of the spring then time period.

0%
Increases
0%
Decreases
0%
Remain same
0%
Time period is independent of I

How much height must an object of mass

$5kg$ be lifted to have potential energy

$5000J$ ?

$[g=10 m/s^2]$

0%
$100m$
0%
$50m$
0%
$120m$
0%
$110m$

Explanation

Gravitation potential energy is given by

$P.E = mgh$

Given:

Mass

$m=5 \ kg$

$g=10 \ m/s^2$

$5000 = 5 \times 10 \times h$

$\implies h = 100 m$

Two particle of masses

$m$ and

$2m$ are colliding elastically as given in figure. If

$V_1$ and

$V_2$ speed of particle just after collision then

0%
$V_1 = 11.16 \,m/s, V_2 = 6.31 \,m/s$
0%
$V_1 = 10.16 \,m/s, V_2 = 5.31 \,m/s$
0%
$V_1 = 9.16 \,m/s, V_2 = 6.31 \,m/s$
0%
$V_1 = 6.31 \,m/s, V_2 = 11.16 \,m/s$

Explanation

Using momentum conservation
In x direction

$m \, x \, 10 \cos 60^o + 2m \dfrac{5}{\sqrt{2}} = \dfrac{mv_2}{\sqrt{2}} + 2 mv_1 \dfrac{\sqrt{3}}{2}$

$5 \sqrt{2} + 10 = v^2 + \sqrt{6} V_1 ...... (A)$
In Y-direction

$2m \dfrac{5}{\sqrt{2}} - m \dfrac{10 \sqrt{3}}{2} = \dfrac{2 mv_1}{2} - \dfrac{mv_1}{\sqrt{2}}$

$10 - 5 \sqrt{6} = \sqrt{2} V_1 - V_2 ... (B)$
Using A and B

$V_1 = 6.31 \, m/s , \, V_2 = 11.16 \, m/s$

A perfectly elastic ball falls on a horizontal floor from a height in a time

$t$ . It will hit the floor again after a time

$t'$ . The ratio of

$t'$ and t is

0%
$1:1$
0%
$1:2$
0%
$2:1$
0%
$1:4$

A man of mass 20 kg is running in a straight line with velocity 10 m/s. What is value of his kinetic energy?

0%
$10\ J$
0%
$100\ J$
0%
$1000\ J$
0%
$10000\ J$

Explanation

The kinetic energy of the particle is given as-

$K.E. = \dfrac{1}{2}mv^2$

$m$ ; mass of the object

$v$ ; speed of the object

$\Rightarrow K.E. = \dfrac{1}{2} \times 20 \times 10^2J$

$\Rightarrow K.E. = 1000 J$

An athelete diving off a high spring board can perform a variety of physical moments in the air before entering the water below. Which one of the following parameters will remain constant during the fall? The athelete's:

0%
linear velocity
0%
linear momentum
0%
moment of inertia
0%
angular momentum

Explanation

Here,

Hovering at all the points one by one,

As a constant force of gravity, always be there on the body,

There will always be change of Linear velocity and Linear momentum,

Also, as the athlete tries different physical moments which makes

its body parts to alter its position from center of mass, resulting in

change of moment of inertia.

And as the gravitational force passes through the center of the Earth, its Torque along the center of Earth will be

$0$ ,

Hence keeping the angular momentum constant, as

$\tau=\dfrac{d\omega}{dt}$

Option

$\textbf D$ is the correct answer

A body of mass

$3\ kg$ is moving with a velocity of

$4\ m/s$ towards left, collides head on with a body of mass

$4\ kg$ moving in opposite direction with a velocity of

$3\ m/s$ . After collision the two bodies stick together and move with a common velocity, which is:

0%
zero
0%
$12\ m/s$ towards left
0%
$12\ m/s$ towards right
0%
$\dfrac{12}{7}m/s$ towards left

Explanation

Given:

$u_{1}=4m/s$

$u_{2}=3m/s$

$m_{1}=3kg$

$m_{2}=4\ kg$

By using the conservation of the meomentum:

$m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v$

$3\times 4+4\times (-3)=(3+4)v$

$v=0$

So option

$A$ is correct.

1523994_1702429_ans_a3c887cba4b24732959c62bb5d53bfc5.png

If a ball is dropped from rest, it bounces from the floor. The coefficient of restitution is

$0.5$ and the speed just before the first bounce is

$5\ m/sec$ . The total time taken by the ball to come to rest is:

0%
$2\ sec$
0%
$1\ sec$
0%
$1.5\ sec$
0%
$0.25\ sec$

A particle of mass

$M$ is moving in a horizontal circle of radius

$R$ with uniform speed

$v$ . When it moves from one point to a diametrically opposite point, its:

0%
momentum does not change
0%
momentum changes by $2Mv$
0%
$KE$ changes by $Mv^{2}$
0%
none of the above

Explanation

Let at one point on the circle the velocity point upwards

Then the momentum

$P=mv$

When the body goes diametrically opposite then the velocity will point downwards, i.e

$-v$

New momentum =

$P'=-mv$

Change in momentum=

$mv-(-mv)=2mv$

A gas expands from

$5\ litre$ to

$205\ litre$ at a constant pressure

$50\ N/m^{2}$ . The work done is

0%
$2000\ J$
0%
$1000\ J$
0%
$10000\ J$
0%
None of these

The block will again collide after time:

0%
$6\ sec$
0%
$4\ sec$
0%
$8\ sec$
0%
$\infty$

A ball of mass

$m_{1}$ is moving with velocity

$v$ . It collides head on elastically with a stationary ball of mass

$m_{2}$ . The velocity of ball becomes

$\dfrac{v}{3}$ after collision, then the value of the ratio

$\dfrac{m_{2}}{m_{1}}$ is:

0%
$1$
0%
$2$
0%
$3$
0%
$4$

A particle , moving horizontally , collides perpendicularly at one end of a rod having equal mass and placed on a smooth horizontal surface

0%
Particle comes to rest if collision is perfectly elastic and center of rod starts to move with the same velocity
0%
particle continues to move along the same direction , whatever is the value of e
0%
particle may get rebound back
0%
velocity of mid-point of the rod will be less than $v_0 / 2$ if the particle gets stuck

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

A body of mass

$8\ kg$ collides elastically with a stationary mass of

$2\ kg$ . If initial

$KE$ of moving mass be

$E$ , the kinetic energy left with it after the collision will be:

0%
$0.80\ E$
0%
$0.64\ E$ `
0%
$0.36\ E$
0%
$0.08\ E$

Percentage of energy of cylinder lost due to collision is:

0%
zero
0%
$50\%$
0%
$66.67\%$
0%
$33.33\%$

Mark the correct statement(s).

0%
The work-energy theorem is valid only for particles
0%
The work-energy theorem is an invariant law of physics
0%
The work-energy theorem is valid only in inertial frames of reference.
0%
The work-energy theorem can be applied in non-inertial frames of reference too.

How many collision are possible between the blocks?

0%
$2$
0%
$4$
0%
$27$
0%
$infinite$

The potential energy

$\psi$ in joule, of a particle of mass

$1\ kg$ , moving in the

$x-y$ plane beys the law

$\psi =3x+4y$ m, where

$(x,y)$ are the coordinates of the particle in metre. If the particle is at rest at

$(6,4)$ at time

$t=0$ , then

0%
The particle has constant acceleration.
0%
The work done by the external forces, the position of rest of the particle and the instant of the particle crossing the $x-$ axis is $25\ J$
0%
The speed of the particle when it crosses the $y-$ axis is $10\ ms^{-1}$
0%
The coordinates of the particle at time $t=4$ are $(-18,-28)$

A car is accelerated on a levelled road and attains a speed of

$5$ times its initial speed. In this process , the kinetic energy of the car:

0%
does not change
0%
becomes $4$ times that of initial kinetic energy
0%
becomes $8$ times that of initial kinetic energy
0%
becomes $25$ times that of initial kinetic energy

Explanation

Lets initial velocity of car is $v$ , final velocity of this car will be $5\ v$

$K.E = \dfrac{1}{2}$ $mv ^{2}$

$\therefore \ K.E\propto v^2$

$K.E_{f}=K.E_i \left(\dfrac{v_f}{v_i}\right)^2$ where

$f$ stand for final and

$i$ for initial.

$K.E_{f}=K.E_i \left(\dfrac{5v}{v}\right)^2=25\ K.E_i$

So, when the speed increases to 5 times of its initial velocity, the kinetic energy will become 25 times.

The correct answer is d) becomes

$25$ times that of initial kinetic energy

A stone is thrown upwards as shown in the diagram. When it reaches

$P,$ which of the following has the greatest value for the stone?

0%
its acceleration
0%
its kinetic energy
0%
its potential energy
0%
its weight

Write true or false for the following statements:
If we know the speed and mass of an object, we can find out its kinetic energy.

0%
True
0%
False

Explanation

Kinetic energy KE of a body of mass m moving with velocity or speed v is given by the formula:

$KE = \dfrac{1}{2} mv^2$
The formula suggests that kinetic energy depends on mass and speed of the object. So, if we know the speed v and mass m of an object, we can find out its kinetic energy.

When a stone of mass 'm' falls through a vertical height 'd', the decrease in the gravitational energy is:

0%
$mg/d$
0%
$mg^2/2$
0%
$mgd$
0%
$md/g$

Explanation

Potential energy stored in a body of mass m and at height 'd' will be given by

$m \times g \times d$
This gravitational potential energy keeps on decreasing with the decrease in height of stone, and when the stone falls completely, this potential energy is lost . So when object loses height

$d$ the corresponding loss in potential Energy is

$= mgd$

Potential energy of a person is minimum when:

0%
Person is standing
0%
Person is sitting on a chair
0%
Person is sitting on the ground
0%
Person is lying on the ground

Explanation

Potential energy of a body is defined as energy of a body due to its position in gravitational field.

In general,
Potential energy

$= M \times g \times h$ where: h is the height above ground.
If the person will be lying on ground then it will have minimum height above the ground therefore potential energy of the person will also be minimum.

An object of mass

$5 \,kg$ falls from a height of

$5 \,m$ above the ground. The loss of potential energy of the mass is:

0%
$250 \,J$
0%
$25 \,J$
0%
$2.5 \,kJ$
0%
$50 \,J$

Explanation

Given,

$Mass = 5 \,kg$

$Height = 5 \,m$
Let

$g = 10 \,ms^{-2}$
Potential energy stored in a body of mass m and height h will be given by

$= m \times g \times h$

$= 5 \,kg \times 10 \,ms^{-2} \times 5 \,m$

$= 250 \,J$
When the mass falls completely, same potential energy is lost

$= 250 \,J$

Write true or false for each statement
A gum bottle lying on a table has no energy.

0%
True
0%
False

Explanation

A gum bottle lying on a table has potential energy by virtue of its position. It is at some height

$h$ above the reference line.

So energy possessed by it =

$P.E.=mgh$

So the statement is false. Answer is option B.

CBSE Questions for Class 11 Medical Physics Work, Energy And Power Quiz 10 - MCQExams.com

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Practice Class 11 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 11 Medical Physics Work, Energy And Power Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Medical Physics Quiz Questions and Answers

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