Physics Test

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Physics Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

A uniform electric field of magnitude 245 V/m is directed in the negative y direction as shown in figure. The coordinates of point P and Q are (- 0.5m, - 0.8m) and (0.3m, 0.7m) respectively. Calculate the potential difference V_Q - V_P along the path shown in the figure,

A
0

B
245V

C
321 V

D
367.5V

Solution
Question 2
1 / -0

A body projected horizontally with a velocity v from a height h has a range R. With what velocity this body has to be projected horizontally from a height h/2 to have the same range?

A
√2v

B
2v

C
6v

D
8v

Solution

In \(y\) direction:
\(h=u t+\frac{1}{2} g t^{2}\)
So, \(t=\sqrt{\frac{2 h}{g}}\)
In \(x\) direction \(R=v \times t\)
Now, In \(y\) direction \(\frac{h}{2}=u T+\frac{1}{2} g T^{2}\)
\(T=\sqrt{\frac{h}{g}}\)
So, comparing ranget \(v \times \sqrt{2 h / g}=v_{1} \times \sqrt{h / g}\)
\(v_{1}=v \sqrt{2}\)
Hence, the correct option is (A)
Question 3
1 / -0

A body starts with velocity u and moves on a straight path with constant acceleration. When its velocity becomes 5u the acceleration is reversed in direction without changing the magnitude. When it returns to starting point its velocity becomes:

A
-u

B
-3u

C
-7u

D
-5u

Solution

When the body is moving forward, \((5 u)^{2}-u^{2}=2 a s\)
\(\Rightarrow 25 u^{2}-u^{2}=2 a s\)
Similarly, when the direction is reversed, after travelling the same distance \(v^{2}-(5 u)^{2}=-2 a s\)
From the two equations, we get \(v=u\)
since the direction is opposite to the first direction, velocity at the starting point is \(-u\)
Hence, the correct option is (A)
Question 4
1 / -0

A body starting with a velocity v returns to its initial position after t second with the same speed along the same line. Acceleration of the particle is :

A
-2v / t

B
2v / t

C
v / 2t

D
t / 2v

Solution

Using Kinematics equations:

−v=v+at

Hence , a=−2v / t

Hence, the correct option is (A)
Question 5
1 / -0

A sharp stone of mass 2 kg falls from a height of 10 m on sand and buries into the sand. It comes to rest in a time of 0.029 second. The depth through which it buries into sand is

A
0.2 m

B
0.15 m

C
0.25 m

D
0.30 m

Solution

Velocity attained just before contact:
\(v=\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 10}=\sqrt{196}=14 \mathrm{m} / \mathrm{s}\)
\(u_{1}=v\)
\(v_{1}=u_{1}+a t\)
\(0=14+a(0.029)\)
\(a=-\left(\frac{14}{0.029}\right)\)
\(=0.029 \times 7\)
\(=0.203 \mathrm{m}\)
\(s=u t+\frac{1}{2} a t^{2}=14(0.029)+\frac{1}{2}\left(\frac{-14}{0.029}\right)(0.029)^{2}\)
\(s=0.029\left(14+\frac{1}{2}(-14)\right)\)
\(=0.029 \times 7\)
\(=0.2 \mathrm{m}\)
Hence, the correct option is (A)
Question 6
1 / -0

When a train starting from rest is uniformly accelerating, a plumb bob hanging from the roof of a compartment is found to be inclined at an angle of 45⁰ with the vertical. The time taken by the train to travel a distance of 1/2 km will be nearly

A
7 S

B
10 S

C
15 S

D
25 S

Solution

Taking, \(g=10 \mathrm{m} / \mathrm{s}^{2}\),
since the bob makes a \(45^{\circ}\) angle with vertical, total force (including pseudo force), and hence in vertical direction equals that of horizontal direction.
\(m g=m a\)
\(a=g=10 \mathrm{m} / \mathrm{s}^{2}\)
\(s=\frac{1}{2} a t^{2}=\frac{1}{2} k m\)
\(10 \mathrm{m} / \mathrm{s}^{2} \times t^{2}=1000 \mathrm{m}\)
\(t^{2}=100\)
\(t=10 s\)
Hence, the correct option is (B)
Question 7
1 / -0

Two bodies of different masses m_a and m_b are dropped from two different heights a and b respectively. Ratio of times taken by the two to drop through these distances is:

A
a : b

B
√b :√a

C
√a :√b

D
a² : b²

Solution

\(s=\frac{1}{2} g t^{2}\)
\(t \propto \sqrt{s}\)
So, \(\frac{t_{1}}{t_{2}}=\frac{\sqrt{s_{1}}}{\sqrt{s_{2}}}=\frac{\sqrt{a}}{\sqrt{b}}\)
Displacement is independent of mass.
Hence, the correct option is (C)
Question 8
1 / -0

A stone dropped from the top of a tower covers 24.5 m in the last second of its fall. Height of the tower is

A
24.5 m

B
44.1 m

C
78.4 m

D
122.5 m

Solution

\(u=0\)
Let total time be \(n\) sec. Distance travelled in last sec, \(\Rightarrow 24.5=\left[\frac{\frac{1}{2} g\left(n^{2}\right)}{\downarrow}-\frac{\frac{1}{2} g(n-1)^{2}}{\downarrow}\right]\)
(dist. travelled in \(n\) secs) (dist. travelled in \((n-1)\) secs) \(24.5=\frac{1}{2} g(2 n-1)\)
\(\Rightarrow 5=2 n-1\)
\(\Rightarrow n=3 \mathrm{sec}\)
So, total height \(=\frac{1}{2} g n^{2}=\frac{1}{2} g(3)^{2}=\frac{9 g}{2}=44.1 \mathrm{m}\)
Hence, the correct option is (B)
Question 9
1 / -0

With two forces acting at a point, the maximum effect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are

A
(2+√2)N and (2−√2)N

B
(2+√3)N and (2−√3)N

C
(2+ 1 / 2 √2)N and (2− 1 / 2 √2)N

D
none of these

Solution

Let \(\mathrm{P}\) and \(\mathrm{Q}\) are forces, we know that \(R=\sqrt{P^{2}+Q^{2}+2 P Q \cos \theta}\) When \(\theta=0^{\circ}, R=4 N=\sqrt{P^{2}+Q^{2}+2 P Q}\)
\(P+Q=4\)
When \(\theta=90^{\circ}, R=3 N\)
\(P^{2}+Q^{2}=9\)
From equation ( 1 ) \((P+Q)^{2}=16 \Rightarrow P^{2}+Q^{2}+2 P Q=16\)
\(\Rightarrow 9+2 P Q=16 \Rightarrow 2 P Q=7\)
Now \((P-Q)^{2}=P^{2}+Q^{2}-2 P Q\)
\(\Rightarrow(P-Q)^{2}=9-7 \Rightarrow P-Q=\sqrt{2}\)
On solving equation
(1) and ( 3 ) \(P=\left(2+\frac{1}{2} \sqrt{2}\right) N\) and \(Q=\left(2-\frac{1}{2} \sqrt{2}\right) N\)
Hence, the correct option is (C)
Question 10
1 / -0

A ball is released from the top of a tower of height h m. It takes T seconds to reach the ground. What is the position of the ball in T / 3 second?

A
h / 9 metres from the ground

B
7h / 9 metres from the ground

C
8h / 9 metres from the ground

D
17h / 18 metres from the ground

Solution

\(h=u t+\frac{1}{2} g t^{2}\)
\(h=\frac{1}{2} g T^{2},\) as \(u=0\)
In \(\frac{T}{3} s\)
\(h_{1}=\frac{1}{2} g\left(\frac{T}{3}\right)^{2}=\frac{1}{9}\left(\frac{1}{2} g T^{2}\right)\)
\(\Rightarrow h_{1}=\frac{h}{9}\) from top.
Hence it is \(=h-h_{1}=h-\frac{h}{9}=\frac{8 h}{9}\) from ground
Hence, the correct option is (C)

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Re-Attempt Weekly Quiz Competition

Physics Test - 44

Result

Physics Test - 44

Score

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SHARING IS CARING

Question 1

0

245V

321 V

367.5V

Solution

Question 2

√2v

2v

6v

8v

Solution

Question 3

-u

-3u

-7u

-5u

Solution

Question 4

-2v / t

2v / t

v / 2t

t / 2v

Solution

Question 5

0.2 m

0.15 m

0.25 m

0.30 m

Solution

Question 6

7 S

10 S

15 S

25 S

Solution

Question 7

a : b

√b :√a

√a :√b

a2 : b2

Solution

Question 8

24.5 m

44.1 m

78.4 m

122.5 m

Solution

Question 9

(2+√2)N and (2−√2)N

(2+√3)N and (2−√3)N

(2+ 1 / 2 √2)N and (2− 1 / 2 √2)N

none of these

Solution

Question 10

h / 9 metres from the ground

7h / 9 metres from the ground

8h / 9 metres from the ground

17h / 18 metres from the ground

Solution

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a² : b²