Physics Test

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Physics Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

A space craft of mass 2000 kg moving with a velocity of 600m/s suddenly explodes into two pieces. One piece of mass 500 kg is left stationary. The velocity of the other part must be

A
600 m/s

B
800 m/s

C
1500 m/s

D
1000 m/s

Solution

Using momentum conservation, $M U=m_{1} v_{1}+m_{2} v_{2}$
Here, $M=2000 \mathrm{Kg}, U=600 \mathrm{m} / \mathrm{s}, m_{1}=500, m_{2}=1500, u_{1}=0 \mathrm{m} / \mathrm{s}$
Substituting these values $M U=m_{1} v_{1}+m_{2} v_{2},$ we get, $2000 \times 600=500 \times 0+1500 \times v_{2}$
$\Rightarrow v_{2}=\frac{2000 \times 600}{1500}=\frac{20 \times 200}{5}=800 \mathrm{m} / \mathrm{s}$
Hence, the correct option is (B)
Question 2
1 / -0

If at the highest point of parabolic path kinetic energy is equal to potential energy, the angle of projection would be:

A
30^o

B
45^o

C
60^o

D
70^o

Solution

Option B is correct At highest point, velocity is only in horizontal direction which is $v_{x}=v \cos \theta$ So, $K E=\frac{1}{2} m v_{x}^{2}$ (at highest point) PE at highest point $=m g h_{\text {max }}$ $=m g \frac{v^{2} \sin ^{2} \theta}{2 g}$
$\frac{1}{2} m v^{2} \cos ^{2} \theta=m g \frac{v^{2} \sin ^{2} \theta}{2 g}$
$\tan \theta=1$
$\Rightarrow \theta=45^{\circ}$
Hence, the correct option is (B)
Question 3
1 / -0

A man and a cart move towards each other. The man weighs 64 kg and the cart 32 kg. The velocity of the man is 5.4 km/hr and that of the cart is 1.8km/hr. When the man approaches the cart, he jumps on to it. The velocity of the cart carrying the man will be:

A
30 km/hr

B
3 km/hr

C
1.8 km/ hr

D
zero

Solution

Using conservation of linear momentum
$m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}$
Thus we get $64 \times 1.5-32 \times 0.5=96 \times v$
$\Rightarrow v=\frac{80}{96} m / s$
$=3 k m / h r$ in the direction of man.
Hence, the correct option is (B)
Question 4
1 / -0

A body projected obliquely with velocity 19.6 m/s has its kinetic energy at the maximum height equal to 3 times its potential energy. Since projection from the ground, its height from ground after 1s is:

(h = maximum height, take g=9.8 m/s²)

A
h/2

B
h/4

C
h/3

D
h

Solution

Kinetic energy at max height $=\frac{1}{2} m(u \cos \theta)^{2}$ Potential energy at max height $=m g \times\left(\frac{u^{2} \sin ^{2} \theta}{2 g}\right)$
$\frac{1}{2} m(u \cos \theta)^{2}=3 m g \times\left(\frac{u^{2} \sin ^{2} \theta}{2 g}\right)$
$\cos ^{2} \theta=3 \times \sin ^{2} \theta$
$\theta=30^{\circ}$
$u=19.6 \mathrm{m} / \mathrm{s}$
After 1 sec $h=u \sin \theta t-\frac{1}{2} g t^{2}$
$h=19.6 \times\left(\frac{1}{2}\right) \times 1-\left(\frac{1}{2}\right) \times 9.8 \times 1^{2}=4.9 m$
$h_{m a x}=\frac{u^{2} \sin ^{2} \theta}{2 g}=19.6^{2} \times\left(\frac{1}{4}\right) \times\left(\frac{1}{2}\right) \times\left(\frac{1}{9.8}\right)=4.9 m=h$
Hence, the correct option is (D)
Question 5
1 / -0

A curve is drawn expressing the kinetic energy of a particle as a function of the distance traversed (on X-axis). The slope of this curve represents the instantaneous:

A
velocity

B
acceleration

C
force

D
power

Solution

K. $E=\frac{1}{2} m v^{2}$
$\frac{d(K \cdot E)}{d x}=m v \cdot \frac{d v}{d x}$
$=m v \frac{d v}{d x} ;$ (substituting $\left.v=\left(\frac{d x}{d t}\right)\right)$
$=m\left(\frac{d x}{d t}\right)\left(\frac{d v}{d x}\right)$ (writing $\left.\left(\frac{d v}{d t}\right)=a\right)$
$=m a=$ force
Hence, the correct option is (C)
Question 6
1 / -0

A bullet is fired from a gun, the force on the bullet is given by F=600−2×10⁵t where F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

A
9 N−s

B
0

C
0.9 N−s

D
1.8 N−s

Solution

Force on the bullet $F=600-2 \times 10^{5} t$
Let the time at which $F=0$ be $t^{\prime}$
$\therefore t^{\prime}=\frac{600}{2 \times 10^{5}}=3 \times 10^{-3} s$
The impulse imparted to bullet
$I=\int_{0}^{t^{\prime}} F d t=\int_{0}^{t^{\prime}}\left(600-2 \times 10^{5} t\right) d t=\left[600 t-10^{5} t^{2}\right]_{0}^{t^{\prime}}$
$\Rightarrow I=600 \times t^{\prime}-10^{5} t^{2}=600 \times 3 \times 10^{-3}-10^{5}\left(3 \times 10^{-3}\right)^{2}=1.8-0.9=0.9 N s$
Hence, the correct option is (C)
Question 7
1 / -0

A bullet fired into a fixed target looses half of its velocity in penetrating 15cm. Before coming to rest, it can penetrate a further distance of:

A
5cm

B
15cm

C
7.5cm

D
10cm

Solution

Assuming it has only $K E$ initially, Let initial velocity be $v$. So, final velocity is $v / 2$ Energy lost is $\frac{1}{2} m v^{2}-\frac{1}{2} m(v / 2)^{2}=\frac{1}{2} m\left(\frac{3}{4} v^{2}\right)$
$$
\begin{array}{r}
=\frac{3}{4}\left(\frac{1}{2} m v^{2}\right) \\
\text { Remaining energy }=\frac{1}{4}\left(1 / 2 m v^{2}\right)
\end{array}
$$
Now $\frac{3}{4} K E$ is lost in $15 \mathrm{cm}$ penetration, hence $\frac{1}{4} K E$ would penetrate to $\frac{15}{3}=5 \mathrm{cm}$ penetration.
Hence, the correct option is (A)
Question 8
1 / -0

A man carries a load of 50kg through a height of 40m in 25seconds. If the power of the man is 1568W, his mass is:

A
5kg

B
1000kg

C
200kg

D
50kg

Solution

Power$=\frac{\text {Work Done}}{t}=\frac{\Delta(P E)}{t}=\frac{m g h}{t}=\frac{\left(50+m_{\text {man}}\right) g \times 40}{25}$
$\Rightarrow 1568=\frac{\left(50+m_{\text {man}}\right) \times 9.8 \times 40}{25}$
$\Rightarrow m_{\text {man}}=50 \mathrm{kg}$
Hence, the correct option is (D)
Question 9
1 / -0

A motor pump set of efficiency 80%, lifts 800 litres of water in 19.6 seconds over a head of 20m. Its input power is (Take g=9.8 m/s²):

A
64kW

B
40kW

C
10kW

D
196kW

Solution

since efficiency of pump is $80 \%,$ we can say, effective power of the pump is $\frac{80 P}{100}$ $m=800 \mathrm{kg}, h=20 \mathrm{m}$ and $g=9.8 \mathrm{m} / \mathrm{s}^{2}$
$t=19.6 s$
Purting these values into $0.8 P=m g h / t,$ we get, $P=10^{4} W=10 \mathrm{kW}$
Hence, the correct option is (C)
Question 10
1 / -0

A tank of size 10m×10m×10m is full of water and built on the ground. If g=10 ms⁻², the potential energy of the water in the tank is:

A
5×10⁷ J

B
1×10⁸ J

C
10⁴ J

D
10⁵ J

Solution

The mass of 1 cubic meter of water $=1000 \mathrm{kg}$ So, the mass of $10 \times 10 \times 10$ cubic meter of water $=10^{6} \mathrm{kg}$ So, $P E=m g h=10^{6} \times 10 \times 5,(5 m$ is the height of the mid point of the tank) So, $P E=5 \times 10^{7} J$
Hence, the correct option is (A)

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Physics Test - 42

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Physics Test - 42

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SHARING IS CARING

Question 1

600 m/s

800 m/s

1500 m/s

1000 m/s

Solution

Question 2

30o

45o

60o

70o

Solution

Question 3

30 km/hr

3 km/hr

1.8 km/ hr

zero

Solution

Question 4

h/2

h/4

h/3

h

Solution

Question 5

velocity

acceleration

force

power

Solution

Question 6

9 N−s

0

0.9 N−s

1.8 N−s

Solution

Question 7

5cm

15cm

7.5cm

10cm

Solution

Question 8

5kg

1000kg

200kg

50kg

Solution

Question 9

64kW

40kW

10kW

196kW

Solution

Question 10

5×107 J

1×108 J

104 J

105 J

Solution

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30^o

45^o

60^o

70^o

5×10⁷ J

1×10⁸ J

10⁴ J

10⁵ J