Gravitation Test

Result

Gravitation Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

If water falls from a tap, then the volume rate of flow of water at depth of $$h,$$ ($$A _ { 0 }$$ is the area of cross-section of the mouth and $$\dfrac { A _ { 0 } } { 2 }$$ is the corresponding area at depth $$h$$ )

A
$${ A }_{ { { 0 } } }\sqrt { 2gh } $$

B
$${ A }_{ { { 0 } } }\sqrt { \dfrac { 2 }{ 3 } gh} $$

C
$$2{ A }_{ 0 }\sqrt { gh } $$

D
$${ A }_{ { { 0 } } }\sqrt { 3gh } $$

Solution
Question 2
1 / -0

Four marbles are dropped from the top of a tower one after the other with an interval of one second. The first one reaches the ground after 4 seconds. When the first one reaches the ground the distances between the first and second, the second and third and the third and fourth will be respectively

A
35, 25 and 15 m

B
30, 20 and 10 m

C
20, 10 and 5 m

D
40, 30 and 20 m

Solution
Question 3
1 / -0

A body dropped freely has covered $${(16/25)}^{th}$$ of the total distance in the last second. Its total time of fall is

A
2.5 s

B
5 s

C
7.5 s

D
1 s
Question 4
1 / -0

In the situation shown , AB is a thin uniform rod of mass m and length l. The gravitation force exerted by it on the particle of m kept at C has magnitude

A
$$\frac { { Gm }^{ 2 } }{ (a+\frac { 1 }{ 2 } { ) }^{ 2 } } $$

B
$$\frac { { Gm }^{ 2 } }{ a(1+a) } $$

C
$$\frac { { Gm }^{ 2 } }{ 1(1+a) } $$

D
$$\frac { { Gm }^{ 2 } }{ 2I(1+a) } $$
Question 5
1 / -0

A body of mass $$10 \ kg$$ is thrown up with a velocity $$98 ms^{-1}$$. Simultaneously another is dropped from a tower. After 2 seconds their momenta are found to be same in magnitude. Then the mass of the freely falling body is

A
40 kg

B
4 kg

C
30 kg

D
20 kg

Solution

Given:
Mass of body 1 is $$m_1=10 \ kg$$
Initial velocity of body 1 is $$u_1=98 \ ms^{-1}$$

Let mass of body 2 be $$m_2.$$
Initial velocity of body 2 is $$u_2=0 \ ms^{-1}$$

Time interval $$t=2 \ sec$$
Let the upward direction be $$(+)ve$$ and downward be $$(-)ve$$.
Acceleration $$a=-g=-9.8 \ m/s^2$$

Final velocity of body 1 is $$v_1.$$
By first equation of motion,
$$v_1=u_1+at$$
$$\Rightarrow v_1=98+(-9.8)(2)$$
$$\Rightarrow v_1=+78.4 \ m/s$$
(positive sign means the velocity is upwards)

Final velocity of body 2 is $$v_2.$$
By first equation of motion,
$$v_2=u_2+at$$
$$\Rightarrow v_2=0+(-9.8)(2)$$
$$\Rightarrow v_2=-19.6 m/s$$
(negative sign means the velocity is downwards)

According to the question, the momentum $$p_1$$ of body 1 and momentum $$p_2$$ of body 2 are equal in magnitude at $$t=2 \sec.$$
$$\therefore \ p_1=p_2$$
$$\Rightarrow m_1v_1=m_2v_2$$
$$\Rightarrow (10)(78.4)=m_2(19.6)$$
$$\Rightarrow m_2=\dfrac{784}{19.6}$$
$$\Rightarrow m_2=40 \ kg$$

Option $$(A)$$ is the answer.
Question 6
1 / -0

A stone is dropped from some height, then the distance travelled by stone in $$4^{th},5^{th},6^{th}$$ seconds are in the ratio of

A
4:5:6

B
16:25:36

C
36:25:16

D
7:9:11
Question 7
1 / -0

A person weighs $$110.84\ N$$ on the moon whose acceleration due to gravity is one-sixth of that earth if the value of g on earth is $$9.8 \ ms^{-2}$$ calculate, acceleration due to gravity on the earth.

A
$$1.89\ ms^{-2}$$

B
$$1.63\ ms^{-2}$$

C
$$2.89\ ms^{-2}$$

D
$$2.63\ ms^{-2}$$

Solution

According to the question acceleration due to gravity on the moon is one sixth of the earth.
Acceleration due to gravity on earth-
$$g_e = 9.8\ ms^{-2}$$
So, acceleration due to gravity on the moon is,
$$g_m = \dfrac{g_e}{6}$$
$$\Rightarrow g_m = \dfrac{9.8}{6}$$
$$\Rightarrow g_m = 1.63\ ms^{-2}$$
Question 8
1 / -0

Two point masses A and B have masses in the ratio of 1 : 4 .The point masses are at distance 1 m apart. At what distance from point mass B. a point mass m is placed so that at that point the resultant force on it zero ?

A
2/3 m

B
1/3 m

C
2/5 m

D
1/5 m

Solution
Question 9
1 / -0

From the top of a tower, a stone is projected vertically upwards with a velocity of $$20 \ m/s$$ and its reaches the ground in $$8 \ sec$$. If $$g=10 \ m/s^2$$, the height of the tower is

A
100m

B
120m

C
160m

D
180m

Solution
Question 10
1 / -0

A body falls from height h=200 m. The ratio of distance travelled in each 2 s, during t=0 to t=6 s of the journey is ($$g= 10 m/s^2$$):

A
1:4:9

B
1:2:4

C
1:3:5

D
1:2:3

Solution