Assertion is incorrect but Reason is correct

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

MCQ Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 15

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 is falling a height

$h$ . After it has fallen a height

$\dfrac{h}{2}$ . it will possess

0%
only potential energy
0%
only kinetic energy
0%
half potential energy and half kinetic energy
0%
more kinetic and less potential energy

Explanation

$(c)$ half potential and half kinetic energy
Explanation: When the body is height

$h$ ; its potential energy is at maximum and kinetic energy is zero. When the body hits the ground, its potential energy becomes zero and kinetic energy is at maximum. At mid-way i.e., half the height; its potential energy becomes half of the maximum potential energy and same happens to the kinetic energy.

Four alternatives are given to each of the following incomplete statements/questions. Choose the right answer:
Which of the following object has higher potential energy?

0%
mass $= 10\,kg$ , $g = 9.8\, ms^{-2} , h = 10\,m$
0%
mass $= 5\,kg , g = 9.8\, ms^{-2} , h = 12\,m$
0%
mass $= 8\,kg , g = 9.8\,ms^{-2} , h = 100\,m$
0%
mass $= 6\,kg , g = 9.8\,ms^{-2} , h = 20\,m$

Explanation

The potential energy is given as-

$PE = mgH$

$m$ ; mass of the object

$g = 9.8/ ms^{-2}$

$h$ ; height gained from ground

For option A:

$m = 10\ kg,\ h = 10\ m$

$PE = 10 \times 9.8 \times 10$

$\Rightarrow PE = 980\ J$

For option B:

$m = 5\ kg,\ h = 12\ m$

$PE = 5 \times 9.8 \times 12$

$\Rightarrow PE = 588\ J$

For option C:

$m = 8\ kg,\ h = 100\ m$

$PE = 8 \times 9.8 \times 100$

$\Rightarrow PE = 7840\ J$

For option D:

$m = 6\ kg,\ h = 20\ m$

$PE = 6 \times 9.8 \times 20$

$\Rightarrow PE = 1176\ J$

For option C potential energy is highest so option C is the correct answer.

A force

$F = K(yi + xj )$ ( where

$K$ is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is

0%
$-2 Ka^2$
0%
$2K a^2$
0%
$-Ka^2$
0%
$Ka^2$

Explanation

While moving from

$(0,0) to (a,0)$

Along positive x-axis

$y = 0 \therefore \overrightarrow {F} = -kx \hat {j}$

i.e. force is in negative y-direction while displacement is in positive x-direction.

$\therefore W_1 = 0$

Because force is perpendicular to displacement

Then particle moves from

$(a,0)$ to

$(a,a)$ along a line parallel to y-axis

$( x = +a)$ during this

$\overrightarrow {F} = -k( y \hat {j} + a \hat {J})$

The first component of force

$- ky \hat {i}$ will not contribute any work because this component is along negative x-direction

$( - \hat {i})$ while displacement is in positive y- direction

$(a,0)$ to

$(a,a)$ The second component of force i.e.

$- ka \hat {j}$ will perform negative work.

$\therefore W_2 = ( - ka \hat {j})(a \hat {j}) = ( -ka)(a) = -ka^2$

So net work done on the particle

$W = W_1 +W_2$

$= 0 + ( -ka^2) =-ka^2$

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

$2v$ .

0%
equal to the work required to accelerate the object from $v = 0$ to $v$ ,
0%
twice the work required to accelerate the object from $v = 0$ to $v$
0%
three times the work required to accelerate the object from $v = 0$ to $v$ ,
0%
four times the work required to accelerate the object from $0$ to $v$
0%
not known without knowledge of the acceleration

Explanation

Answer (c). The net work needed to accelerate the object from

$v = 0$ to

$v$ is

$W_{1}=KE_{1f}-KE_{1i}=\dfrac{1}{2}mv^{2}-\dfrac{1}{2}m(0)^{2}=\dfrac{1}{2}mv^{2}$
The work required to accelerate the object from speed

$v$ to speed

$2v$ is

$W_{2}=KE_{2f}-KE_{2i}=\dfrac{1}{2}m(2v)^{2}-\dfrac{1}{2}mv^{2}$

$=\dfrac{1}{2}m(4v^{2}-v^{2})=3\left ( \dfrac{1}{2}mv^{2} \right )=3W_{1}$

Bullet

$2$ has twice the mass of bullet

$1$ . Both are fired so that they have the same speed. If the kinetic energy of bullet

$1$ is

$K$ , is the kinetic energy of bullet

$2$

0%
$0.25K$
0%
$0.5K$
0%
$0.71K$
0%
$K$
0%
$2K$

Explanation

Let

$m$ be the mass of bullet

$1$ .

According to the given condition, the mass of bullet

$2$ is

$2m.$

Kinetic energy of bullet

$1$ is

$KE_1=\dfrac{1}{2}mv^2=K$

KE of bullet

$2$ is

$KE_2=\dfrac{1}{2}(2m)v^2=2(\dfrac{1}{2}mv^2)=2K$

The displacement of a particle of mass

$m=2\,kg$ on a smooth horizontal surface is a function of time given by

$x=e^{2t}$ . Find net the work done by external agent from

$t=\,ln\,1$ to

$t=\,ln\,2$ .

0%
$20\,J$
0%
$30\,J$
0%
$50\,J$
0%
$60\,J$

Explanation

Displacement of particle

$x=e^{2t}$

$\therefore$ velocity at any instant

$'t'$ is given by

$v=\dfrac{dx}{dt}=2e^{2t}$

$\therefore$ acceleration at any instant

$'t'$ is given by

$a=\dfrac{dv}{dt}=4e^{2t}=\dfrac{d^2x}{dt^2}$

work done by the force

$\displaystyle W=\int F.dx=\int F.\dfrac{dx}{dt}.dt$

$\displaystyle W= \int^{ln\,2}_{ln\,1}ma\left(\dfrac{dx}{dt}\right)dt$

$m=2kg$

$\displaystyle =\int^{ln\,2}_{ln\,1}2\times 4e^{2t}\times 2e^{2t}dt$

$\displaystyle =16\int^{ln\,2}_{ln\,1}e^{4t}dt=\dfrac{16}{4}[e^{4t}]^{ln\,2}_{ln\,1}$

$=\dfrac{16}{4}[e^{4\,ln\,2}-e^{4\,ln\,1}]$

$=4[e^{ln\,(2^4)}-e^{ln\,(1^4)}]$

$=4[2^4-1^4]=4[15]=60\,Joules$

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 15 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 15 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.