Work Power & Energy — Page 10 | NEET Physics

Q91

A ball of mass

0.2

kg

is thrown vertically upwards by applying a force by hand. If the hand moves

0.2

m

while applying the force and the ball goes upto

2

m

height further, find the magnitude of the force. (consider

g = 10\,m/{s^2}

).

A

4N

B

16

N

C

20

N

D

22

N

Correct Answer

Option D

Solution

According to energy conservation law, Work done by the hand and due to gravity = total change in the kinetic energy Initially the the ball is at rest and finally at top its velocity become zero so total change in kinetic energy

\Delta K

= 0

{W_{hand}} + {W_{gravity}} = \Delta K

[Here distance covered would be 0.2 meter for force by hand as force is applied while ball is in contact with hand.

And gravity will still work while ball is in contact with hand so total distance due to gravity would be 2 + 0.2 = 2.2 meter.]

\Rightarrow F\left( {0.2} \right) - \left( {0.2} \right)\left( {10} \right)\left( {2.2} \right)

= 0 \Rightarrow F = 22\,N

$\therefore$ Option (D) is correct.

Q92

A force

\overrightarrow{\mathrm{F}}=2 \hat{i}+\mathrm{b} \hat{j}+\hat{k}

is applied on a particle and it undergoes a displacement

\hat{i}-2 \hat{j}-\hat{k}

What will be the value of

b

, if work done on the particle is zero.

A

\dfrac{1}{3}

B

\dfrac{1}{2}

C 0

D 2

Correct Answer

Option B

Solution

To determine the value of

b

such that the work done on the particle is zero, follow these steps: The work done by a force on a displacement is given by the dot product:

\text{Work} = \overrightarrow{F} \cdot \overrightarrow{d}.

Given the force

\overrightarrow{F} = 2\hat{i} + b\hat{j} + \hat{k},

and the displacement

\overrightarrow{d} = \hat{i} - 2\hat{j} - \hat{k},

compute the dot product:

\overrightarrow{F} \cdot \overrightarrow{d} = (2)(1) + (b)(-2) + (1)(-1).

Simplify the expression:

\overrightarrow{F} \cdot \overrightarrow{d} = 2 - 2b - 1 = 1 - 2b.

Since the work done is zero:

1 - 2b = 0.

Solve for

b

:

2b = 1,

b = \frac{1}{2}.

Thus, the value of

b

is

\frac{1}{2}

, which corresponds to Option B.

Q93

A sand dropper drops sand of mass m(t) on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e.,

\dfrac{dm}{dt} \propto \sqrt{v}

. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?

A P

\propto \sqrt{v}

B P

\propto v

C

P^2 \propto v^3

D

P^2 \propto v^5

Correct Answer

Option D

Solution

Given,

{{dm} \over {dt}} \propto \sqrt v

So,

{{dm} \over {dt}} = k\sqrt v

.... (i) where, k = constant We know, Power =

P = FV

\Rightarrow P = {{dp} \over {dt}}v

(As

F = {{dp} \over {dt}}

where, p = linear momentum)

\Rightarrow P = {d \over {dt}}(mv)v

\Rightarrow P = v{{dm} \over {dt}}\,.\,v

(As speed is constant i.e. v = constant)

\Rightarrow P = {v^2}(k\sqrt v )

[From (i)]

\Rightarrow P = k{v^{5/2}}

by squaring both sides,

\Rightarrow {P^2} = {k^2}{v^5}

\Rightarrow {P^2} \propto {v^5}

(As $\mathrm{k^2}$ = constant) Hence, option 4 is correct.

Q94

A force

\mathrm{F}=\alpha+\beta \mathrm{x}^2

acts on an object in the x -direction. The work done by the force is 5 J when the object is displaced by 1 m . If the constant

\alpha=1 \mathrm{~N}

then

\beta

will be

A

15 \mathrm{~N} / \mathrm{m}^2

B

10 \mathrm{~N} / \mathrm{m}^2

C

12 \mathrm{~N} / \mathrm{m}^2

D

8 \mathrm{~N} / \mathrm{m}^2

Correct Answer

Option C

Solution

The work done by a force is given by the integral of the force over the displacement. The force is given by:

F(x) = \alpha + \beta x^2

To find the work done when the object is displaced from $x = 0$ to $x = 1$ , we compute:

W = \int_{0}^{1} (\alpha + \beta x^2) \, dx

Substituting $\alpha = 1 \, \text{N}$ :

W = \int_{0}^{1} (1 + \beta x^2) \, dx = \int_{0}^{1} 1 \, dx + \int_{0}^{1} \beta x^2 \, dx

Calculating each integral separately: $\int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1$ $\int_{0}^{1} \beta x^2 \, dx = \beta \left[ \dfrac{x^3}{3} \right]_{0}^{1} = \beta \left[\dfrac{1^3}{3} - \dfrac{0^3}{3}\right] = \dfrac{\beta}{3}$ Thus, the total work done is:

W = 1 + \frac{\beta}{3}

We are given that the work done is 5 J, so:

1 + \frac{\beta}{3} = 5

Subtract 1 from both sides:

\frac{\beta}{3} = 4

Multiply both sides by 3 to solve for $\beta$ :

\beta = 12 \, \text{N/m}^2

Therefore, the value of $\beta$ is $\boxed{12 \, \text{N/m}^2}$ .

Thus, Option C $12 \, \text{N/m}^2$ is the correct answer.

Q95

Two bodies are having kinetic energies in the ratio 16 : 9. If they have same linear momentum, the ratio of their masses respectively is :

A

3: 4

B

4: 3

C

9: 16

D

16: 9

Correct Answer

Option C

Solution

The kinetic energy of a body of mass $m$ and velocity $v$ is given by $K=\dfrac{1}{2}mv^2$ .

Since the bodies have the same linear momentum, we can write:

p=mv

where $p$ is the linear momentum of the bodies.

Let the masses of the two bodies be $m_1$ and $m_2$ and their kinetic energies be $K_1$ and $K_2$ , respectively.

Then, we have:

\frac{K_1}{K_2}=\frac{16}{9}

\frac{1}{2}m_1v_1^2\div\frac{1}{2}m_2v_2^2=\frac{16}{9}

Since $p=mv$ , we have $v_1=\dfrac{p}{m_1}$ and $v_2=\dfrac{p}{m_2}$ . Substituting these in the above equation, we get:

\frac{m_2}{m_1}=\frac{9}{16}

Therefore, the ratio of the masses of the two bodies is $\boxed{9:16}$ .

Q96

A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by n2 R t2 where n is a constant. The power delivered to the particle by the force acting on it, is :

A M n2 R2 t

B M n R2 t

C M n R2 t2

D

{1 \over 2}

M n2 R2 t2

Correct Answer

Option A

Solution

We know, centripetal acceleration =

{{{V^2}} \over R}

$\therefore$ According to question,

{{{V^2}} \over R}

=

{n^2}R{t^2}

$\Rightarrow$ V2 = n2 R2 t2 $\Rightarrow$ V = nRt $\Rightarrow$

{{dV} \over {dt}}

= nR Power (P) = Force (F) $\times$ Velocity (V) = M

{{dV} \over {dt}}

(V) = M (nR) (nRt) = Mn2R2t

Q97

A wire suspended vertically from one of its ends is stretched by attaching a weight of

200N

to the lower end. The weight stretches the wire by

1

mm.

Then the elastic energy stored in the wire is

A

0.2

J

B

10

J

C

20

J

D

0.1

J

Correct Answer

Option D

Solution

The elastic potential energy

= {1 \over 2} \times

Force $\times$ extension

= {1 \over 2} \times 200 \times 0.001 = 0.1\,J