Motion in A Straight Line Test

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Motion in A Straight Line Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

An ideal spring having force constant k is suspended from the rigid support and a block of mass M is attached to its lower end. The mass is released from the natural length of the spring. Then the maximum extension in the spring is

A
4Mg/k

B
2Mg/k

C
Mg/k

D
Mg/2k

Solution

Question 2

1 / -0

The physical situation in List I with graphs of the variation of total energy ($$E$$), potential energy ($$U$$) and kinetic energy ($$K$$) with time in List II are given. Match the statements from List I with those in List II and select the correct answer using the code given below the lists.

List I	List II
(P) A mass attached to an unstretched ideal spring, released from position O in vertical plane from rest until it reaches its maximum extension. Assume that the gravitational P.E reference is at the equilibrium position of $$m$$
(Q) An object of mass is released from tower AB undergoes free fall
(R) An object of mass $$m$$ is being pulled by a constant force in horizontal direction on a horizontal frictionless surface
(S) A particle of mass $$m$$ is given velocity $${v}_{0}$$ downward on a rough inclined surface whose coefficient of friction is $$\mu=\tan{\theta}$$

P- 1; Q- 4; R- 2; S- 3

P- 2; Q- 3; R- 1; S- 4

P- 4; Q- 1; R- 2; S- 3

P- 4; Q- 1; R- 3; S- 2

Solution

Question 3
1 / -0

The pulley and string are shown in Fig. smooth and of negligible mass. For the system to remain in equilibrium, the angle $$\theta$$ should be

A
$$0^0$$

B
$$30^0$$

C
$$45^0$$

D
$$60^0$$

Solution

We know that,
Tension on the string $$=mg$$ $$\longrightarrow \left( i \right) $$
The component of $$T$$ which is working $$\Rightarrow T\cos\theta $$
also, $$T\cos\theta =\sqrt { 2 } mg$$
Since there are $$2$$ blocks,
$$2T\cos\theta =\sqrt { 2 } mg$$
$$2\left( mg \right) \cos\theta =\sqrt { 2 } mg$$ (from (i))
$$\cos\theta =\dfrac { 1 }{ \sqrt { 2 } } $$
$$\theta ={ \cos }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } } $$
[C] $$\boxed { \theta ={ 45 }^{ 0 } } $$
Question 4
1 / -0

A swimmer is capable of swimming $$1.65$$ $$ms^{-1}$$ in still water. If she swims directly across a $$180$$m wide river whose current is $$0.85$$ m/s, how far downstream(from a point opposite her standing point) will she reach?

A
$$92.7$$ m

B
$$40$$ m

C
$$48$$ m

D
$$20$$ m

Solution

Time taken to cross the river $$t = \dfrac{180}{1.65} = 109 \ s$$
Velocity of river $$V_R = 0.85 \ m/s$$
Drift $$x =V_R\times t = 0.85\times 109 = 92.7 \ m $$
Question 5
1 / -0

Directions For Questions

A car accelerates from rest constant rate of $$2\ {ms}^{-2}$$ for some time. Immediately after this, it retards at a constant rate of $$4\ {ms}^{-2}$$ and comes to rest. The total time for which it remains in motion is $$3\ s$$. Taking the moment of start of motion as $$t=0,$$ answer the following question.

...view full instructions

Let $$t$$ be a time instant greater than $${t}_{1}$$ but less than $$3\ s$$. The velocity at this time instant is :

A
$$4\ (t-{t}_{1})$$

B
$$2\ {t}_{1}+4\ (t-{t}_{1})$$

C
$$2\ {t}_{1}-4\ (t-{t}_{1})$$

D
$$4t$$

Solution
Question 6
1 / -0

During a rainy day, rain is falling vertically with a velocity $$2\ ms^{-1}$$. A boy at rest starts his motion with a constant acceleration of $$2ms^{-2}$$ along a straight road. Then the rate at which the angle of the axis of umbrella with the vertical should be changed so that the rain always appears to fall parallel to the axis of umbrella is

A
$$\dfrac{4}{4+t^{2}}$$

B
$$\dfrac{3}{3+t^{2}}$$

C
$$\dfrac{2}{2+t^{2}}$$

D
$$\dfrac{1}{1+t^{2}}$$

Solution

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Question 7
1 / -0

A traveller while in a uniformly moving train throws a ball up in the air. The ball will return-

A
In his hand

B
Ahead in the direction of motion of the train

C
Trail behind

D
Deflected sideways

Solution

As the traveller is moving in horizontal direction with uniform velocity
Even the ball will gain initial horizontal velocity and there is no acceleration in horizonal direction so the ball continues to move with same horizontal velocity
So it returns back to his hand
Question 8
1 / -0

A person is travelling from the ground floor to the first floor in a mall using an escalator. Consider the following three separate cases. When the person is standing on the moving escalator it takes one minute for him to reach the first floor. If the escalator does not move it takes him $$3$$ minutes to walk up the stationary escalator to reach the top. How long will it take for the person to reach the top if he walks up the moving escalator?

A
$$30\ \sec$$

B
$$45\ \sec$$

C
$$40\ \sec$$

D
$$35\ \sec$$

Solution

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Question 9
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A stone falls from rest. The total distance covered by it in the last second of its motion is equal to the distance covered in the first three seconds. What is the height from which the stone was dropped? Take g = 10 $$m/s^2$$:

A
45 m

B
100 m

C
125 m

D
200 m

Solution

Distance covered in 3 sec$$=\dfrac{1}{2}gt^2=\dfrac{1}{2}\times10\times 3^2=45m$$
At the end of 'n' second its velocity $$u=gn$$
At the end of 'n+1' second its velocity $$v=g(n+1)$$

Distance traveled in the (n+1)th second$$=\dfrac{(2n+1)g}{2}=\dfrac{(20n+10)}{2}=10n+5$$

$$10n+5=45\Rightarrow (n+1)=5$$
Distance covered in 5 second $$=\dfrac{1}{2}gt^2=\dfrac{1}{2}\times10\times 5^2=125m$$
Question 10
1 / -0

Two blocks $$(A)\ 2\ kg$$ and $$(A)\ 5\ kg$$ rest one over the other on a smooth horizontal plane. The coefficient of static and dynamic friction between $$(A)$$ and $$(B)$$ is the same and equal to $$0.60$$. The maximum horizontal force $$F$$ that can be applied to $$(B)$$ in order that both $$(A)$$ and $$(B)$$ do not have any relative motion is :

A
$$42\ N$$

B
$$84\ kgf$$

C
$$5.4\ kgf$$

D
$$1.2\ N$$

Solution

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