Work and Energy Test

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Work and Energy Test - 54

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Question 1
1 / -0

An object is dropped from height $$h=2R$$ on the surface of earth. Find the speed with which it will collide with ground, by neglecting effect of air. [$$R$$ is radius of earth. Take mass of earth as $$M$$]

A
$$2 \sqrt {\dfrac {GM} {3R}}$$

B
$$\sqrt {\dfrac {2GM} {R}}$$

C
$$\sqrt {\dfrac {GM} {R}}$$

D
$$\sqrt {\dfrac {GM} {2}}$$

Solution
Question 2
1 / -0

The mass of object X is $$M_1$$ and that of object Y is $$M_2$$. Keeping their kinetic energy constant, if the velocity of object Y is doubled the velocity of object X, what will be the relation between their masses?

A
$$M_1=2M_2$$

B
$$4M_1=M_2$$

C
$$M_1=4M_2$$

D
$$2M_1=M_2$$

Solution

Mass of object X = $$M_1$$ , Velocity of object X = $$v_1$$
Mass of object Y = $$M_2$$, Velocity of object Y = $$v_2$$
Also, $$v_2$$ = 2$$v_1$$
K.E is constant.

K.E$$_1$$ = K.E$$_2$$
$$\dfrac{1}{2}$$ $$M_1$$ $$v_1^2$$ = $$\dfrac{1}{2}$$ $$M_2$$ $$v_2^2$$
$$\dfrac{1}{2}$$ $$M_1$$ $$v_1^2$$ = $$\dfrac{1}{2}$$ $$M_2$$ $$(2v_1)^2$$

$$\dfrac{1}{2}$$ $$M_1$$ $$v_1^2$$ = $$\dfrac{1}{2}$$ $$M_2$$ $$4\ v_1^2$$

$$M_1$$ = 4 $$M_2$$
Question 3
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A car weighing 1 ton is moving twice as fast as another car weighing 2 ton. The kinetic energy of the one-ton car is

A
less than that of the two-ton car is

B
some as that of the two-ton car is

C
more than that of the two-ton car is

D
impossible to compare with that of the two-ton car unless the height of each

Solution

$$K_1=\dfrac{1}{2}(1)(2v)^2=2v^2$$
$$K_2=\dfrac{1}{2}(2)(v)^2=v^2$$

$$\implies K_1 > K_2 $$
Question 4
1 / -0

The work function of tungsten is $$4.50eV$$. The wavelength of fastest electron emitted when light whose photon energy is $$5.50eV$$ falls on tungsten surface, is

A
$$12.27\mathring {A}$$

B
$$0.286\mathring {A}$$

C
$$12400\mathring{A}$$

D
$$1.227\mathring{A}$$

Solution
Question 5
1 / -0

Two riffles fire the same number of bullets in a given interval of time. The second fires bullets of mass twice that fired by the first and with a velocity that is half that of the first. The ratio of their powers is?

A
$$1:4$$

B
$$4:1$$

C
$$1:2$$

D
$$2:1$$

Solution

$$\textbf {Hint :}$$ Use $$Power=\dfrac{Energy}{time}$$

$$\textbf{Step 1: Calculating power of first riffle}$$
Let the mass of bullet be $$m$$ and velocity be $$v$$
Power of first riffle $$P_{1}$$ $$\dfrac{Energy}{time}=\dfrac{\frac{1}{2}mv^{2}}{t}........(1)$$
$$\textbf{Step 2: Calculating power of second riffle}$$
Given mass becomes $$2m$$ and velocity becomes $$\dfrac{v}{2}$$
Power of second riffle $$P_2$$ $$=\dfrac { \frac { 1 }{ 2 } \times 2m\times \dfrac { v^{ 2 } }{ 4 } }{ t } ........(2)$$
$$\textbf{Step 3: Ratio}$$
Equating equation (1) and (2)
$$\Rightarrow \dfrac { P_{ 1 } }{ P_{ 2 } } =2:1$$

$${\textbf{Correct option: D}}$$
Question 6
1 / -0

Refer above figure.
An electric bulb 'P' rated 220 V, 60 W and, another identical bulb 'Q' is connected across the mains as shown here.
The power consumed is _ _ _ _ _

A
20W

B
40W

C
30W

D
60W

Solution

Given: Rating of $$P-220V,60W$$

Also, resistance of P and Q are identical
$$P=\dfrac{V^2}{R}$$

$$\boxed{R=\dfrac{V^2}{P}=\dfrac{220\times 220}{60}}$$...............(1)

Now, total resistance $$R+R=2R$$
$$=\dfrac{220\times 220}{60}\times 2$$

$$V=IR_{net}$$ $$I=current$$
$$\therefore$$ Current = $$\dfrac{V}{R_{net}}=\dfrac{220\times 60}{220\times 220\times 2}=\dfrac{30}{220}$$

$$\Rightarrow $$ Current (I) $$=\dfrac{3}{22}$$

$$\therefore Power=I^2R_{net}=\dfrac{3}{22}\times \dfrac{220\times 220\times 2}{60}\times \dfrac{3}{22}=30W$$

Option C is correct.
Question 7
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A pump of $$200W$$ power is lifting $$2kg$$ water from an average depth of $$10m$$ in one second. Velocity of water delivered by the pump is :
$$(g=10m/s^2)$$

A
$$10m/s$$

B
$$2m/s$$

C
$$4 m/s$$

D
$$1 m/s$$

Solution

acceleration due to gravity $$g = 10m/sec^2$$
height $$h = 10\ meters$$
mass $$= 2kg$$
Potential energy $$= mgh = 2\times 10 \times 10 = 200J$$
as power $$= \dfrac{work \ done}{time}$$
when power of motor $$= 200w$$
$$200 = 200/t$$
$$\Rightarrow t = 1sec$$
here displacement of water= height $$= 10m$$
time = 1sec
Hence,
Velocity $$V= \dfrac{Displacement}{Time}$$
$$V= \dfrac{10}{1}$$
$$V= 10m/sec$$
Question 8
1 / -0

A man raises 1 kg wt. to a height of 100 cm and holds it there for 30 minutes. How much work has he performed?

A
$$1\times 9.8J$$

B
$$1\times 9.8\times 30\times 60J$$

C
$$1\times 9.8\times 30J$$

D
$$1\times 9.8\times 30erg$$

Solution

Given
$$m=1\ kg\\h=100\ cm=1\ m \\t=30\ mit$$

Work dont $$W=$$ Force $$\cdot $$ Displacement
Force apply to lift the mass = Weight of the body= $$1\times 9.8$$ N
From above equation,
$$W=1\times 9.8\times 1\ J$$
$$W=1\times 9.8 \ J$$

Option A
Question 9
1 / -0

A skater of weight $$30 \ kg$$ has initial speed $$32m/s$$ and second one of weight $$40 \ kg$$ has $$5m/s$$. After the collision, they stick together and have a speed $$5m/s$$. Then the loss in KE is

A
$$48J$$

B
$$96J$$

C
Zero

D
None of these

Solution

Given:
Weight of the skater 1 is $$m_1=30 \ kg$$
His initial speed $$u_1=32 \ m/s$$
Weight of the skater 2 is $$m_2=40 \ kg$$
His initial speed $$u_2=5 \ m/s$$
After collision they stick together
Both of their speed $$v_1=v_2=v=5 \ m/s$$

We know that kinetic energy $$=\dfrac{1}{2}mv^2$$
Initial kinetic energy $$=KE_{skater 1}+KE_{skater 2}$$
$$=\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2$$
$$=\dfrac{1}{2}(30)(32)^2+\dfrac{1}{2}(40)(5)^2$$
$$=\dfrac{1}{2}(30)(1024)+\dfrac{1}{2}(40)(25)$$
$$=15360+500=15860 \ J$$
Final kinetic energy $$=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2$$
$$=\dfrac{1}{2}m_1v^2+\dfrac{1}{2}m_2v^2=\dfrac{1}{2}(m_1+m_2)v^2$$
$$=\dfrac{1}{2}(30+40)(5)^2$$
$$=\dfrac{1}{2}(70)(25)=875 \ J$$

Loss in kinetic energy $$\Delta KE=15860-875$$
$$=14985 \ J$$
The answer is none of the given options. So option D is the answer.
Question 10
1 / -0

Two bodies A and B have masses $$20\ kg$$ and $$5\ kg$$ respectively . Each one is acted upon by a force of $$4\ kg.-wt$$. If they acquire the same kinetic energy in times $$t_{ A }\ and\ t_{ B }$$, then the ratio $$\dfrac { t_{ A } }{ t_{ B } } $$:

A
$$\dfrac { 1 }{ 2 } $$

B
$$2$$

C
$$\dfrac { 2 }{ 5 } $$

D
$$\dfrac { 5 }{ 6 } $$

Solution