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Laws of Motion

NEET Physics · 100 questions · Page 9 of 10 · Click an option or "Show Solution" to reveal answer

Q81
Three forces F1=10 N,F2=8 N, F3=6 NF_{1}=10 \mathrm{~N}, F_{2}=8 \mathrm{~N}, \mathrm{~F}_{3}=6 \mathrm{~N} are acting on a particle of mass 5 kg5 \mathrm{~kg}. The forces F2\mathrm{F}_{2} and F3\mathrm{F}_{3} are applied perpendicularly so that particle remains at rest. If the force F1F_{1} is removed, then the acceleration of the particle is:
A 4.8 ms−24.8 \mathrm{~ms}^{-2}
B 7 ms−27 \mathrm{~ms}^{-2}
C 2 ms−22 \mathrm{~ms}^{-2}
D 0.5 ms−20.5 \mathrm{~ms}^{-2}
Correct Answer
Option C
Solution

Since the particle is initially at rest, the net force acting on it is zero. This means that the forces

F1F_1

,

F2F_2

, and

F3F_3

are balanced. Given that

F2F_2

and

F3F_3

are acting perpendicularly, we can represent the balance of forces as follows:

F1=F22+F32F_1 = \sqrt{F_2^2 + F_3^2}

We can now calculate the equivalent force resulting from

F2F_2

and

F3F_3

:

F1=(8 N)2+(6 N)2=64+36=100=10 NF_1 = \sqrt{(8 \mathrm{~N})^2 + (6 \mathrm{~N})^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \mathrm{~N}

Since the particle is initially at rest, when the force

F1F_1

is removed, only

F2F_2

and

F3F_3

remain. The net force acting on the particle is the equivalent force resulting from

F2F_2

and

F3F_3

, which we have calculated to be

10 N10 \mathrm{~N}

. Now, we can use Newton's second law to find the acceleration of the particle:

F=maF = ma

Where:

FF

is the net force acting on the particle

mm

is the mass of the particle

aa

is the acceleration of the particle Rearranging the equation to solve for

aa

:

a=Fma = \frac{F}{m}

Plugging in the values for

FF

and

mm

:

a=10 N5 kg=2 ms−2a = \frac{10 \mathrm{~N}}{5 \mathrm{~kg}} = 2 \mathrm{~ms}^{-2}

The acceleration of the particle when the force

F1F_1

is removed is

2 ms−22 \mathrm{~ms}^{-2}

, which corresponds to Option C.

Q82
The initial mass of a rocket is 1000 kg. Calculate at what rate the fuel should be burnt so that the rocket is given an acceleration of 20 ms-2. The gases come out at a relative speed of 500 ms−-1 with respect to the rocket : [Use g = 10 m/s2]
A 6.0 ×\times 102 kg s−-1
B 500 kg s−-1
C 10 kg s−-1
D 60 kg s−-1
Correct Answer
Option D
Solution
Fthrust=(dmdt.Vrel){F_{thrust}} = \left( {{{dm} \over {dt}}.{V_{rel}}} \right)
(dmdtVrel−mg)=ma\left( {{{dm} \over {dt}}{V_{rel}} - mg} \right) = ma
⇒(dmdt)×500−103×10=103×20\Rightarrow \left( {{{dm} \over {dt}}} \right) \times 500 - {10^3} \times 10 = {10^3} \times 20
dmdt{{dm} \over {dt}}

= (60 kg /s)

Q83
A heavy iron bar, of weight WW is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle θ\theta with the horizontal. The weight experienced by the person is :
A Wsin⁡θW \sin \theta
B WW
C W2\dfrac{W}{2}
D Wcos⁡θW \cos \theta
Correct Answer
Option C
Solution
mg×L2cos⁡θ=N×Lcos⁡θ⇒N=mg2=W2\begin{aligned} & m g \times \frac{L}{2} \cos \theta=N \times L \cos \theta \\ & \Rightarrow \quad N=\frac{m g}{2}=\frac{W}{2} \end{aligned}
Q84
A marble block of mass 22 kgkg lying on ice when given a velocity of 66 m/sm/s is stopped by friction in 1010 s.s. Then the coefficient of friction is
A 0.020.02
B 0.030.03
C 0.040.04
D 0.060.06
Correct Answer
Option D
Solution

The retarding force is created by the frictional force. Given,

u=6m/s, v=0, t=10s,u = 6m/s,\,v = 0,\,t = 10s,

Retardation

(a)=−fm=−umgm=−μg=−10μ(a) = - {f \over m} = {{ - umg} \over m} = - \mu g = - 10\mu
v=u+atv = u + at
0=6−10μ×100 = 6 - 10\mu \times 10

∴\therefore

μ=0.06\mu = 0.06
Q85
The position vector of a particle related to time tt is given by r⃗=(10ti^+15t2j^+7k^)m\vec{r}=\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right) m The direction of net force experienced by the particle is :
A Positive xx - axis
B Positive yy - axis
C Positive zz - axis
D In xx - yy plane
Correct Answer
Option B
Solution

To find the direction of the net force experienced by the particle, we need to find the acceleration vector of the particle and then use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration vector.

The position vector of the particle is given by:

r⃗=(10ti^+15t2j^+7k^) m\vec{r} = (10t\hat{i} + 15t^2\hat{j} + 7\hat{k})\,\text{m}

Differentiating r⃗\vec{r} twice with respect to time tt, we get the acceleration vector:

a⃗=d2r⃗dt2=ddt(10i^+30tj^)=30j^ m/s2\vec{a} = \frac{d^2\vec{r}}{dt^2} = \frac{d}{dt}(10\hat{i} + 30t\hat{j}) = 30\hat{j}\,\text{m/s}^2

Therefore, the acceleration vector is a⃗=30j^ m/s2\vec{a} = 30\hat{j}\,\text{m/s}^2.

Using Newton's second law, the net force on the particle is given by:

F⃗net=ma⃗\vec{F}_{net} = m\vec{a}

where mm is the mass of the particle.

Since we are only interested in the direction of the net force, we can ignore the magnitude of the acceleration and focus on its direction, which is along the positive yy-axis.

Q86
A lift is moving down with acceleration a.a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively
A g,gg,g
B g−a,g−ag-a, g-a
C g−a,gg-a, g
D a,ga, g
Correct Answer
Option C
Solution

Let acceleration of ball =

a→b{\overrightarrow a _b}

and acceleration of man is =

a→m{\overrightarrow a _m}

With respect to the man standing in the lift, the acceleration of the ball

a→bm=a→b−a→m{\overrightarrow a _{bm}} = {\overrightarrow a _b} - {\overrightarrow a _m}
⇒abm=g−a\Rightarrow {a_{bm}} = g - a

Where

aa

is the acceleration of the man as the acceleration of the lift is

aa

. With respect to the man standing on the ground the acceleration of the ball

a→bm=a→b−a→m{\overrightarrow a _{bm}} = {\overrightarrow a _b} - {\overrightarrow a _m}
⇒abm=g−0\Rightarrow {a_{bm}} = g - 0
=g= g
Q87
Given below are two statements : Statement I : An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable. Statement II : Force exerted by the floor of an elevator on the foot of a person standing on it is more than his/her weight when the elevator goes down with increasing speed. In the light of the above statements, choose the correct answer from the options given below :
A Both Statement I and Statement II are false
B Both Statement I and Statement II are true
C Statement I is false but Statement II is true
D Statement I is true but Statement II is false
Correct Answer
Option D
Solution

Statement I says that when the weight of an elevator is balanced with the tension of its cable, it can move up or down with a uniform speed.

This is true because the weight of the elevator is balanced by the tension in the cable, which allows it to move smoothly and at a constant speed.

Statement II says that the force exerted by the floor of an elevator on a person's foot is greater than their weight when the elevator goes down with increasing speed.

This is false because the force exerted by the floor on a person's foot is equal to their weight, regardless of the speed of the elevator.

The person's weight is a constant force and does not change with the speed of the elevator.

The apparent weight of a person may change with the speed of the elevator, but this is due to the effects of acceleration and not an increase in the force exerted by the floor.

Q88
A bullet of mass 20 g has an initial speed of 1 ms–1 , just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of 2.5 × 10–2 N, the speed of the bullet after emerging from the other side of the wall is close to :
A 0.3 ms-1
B 0.1 ms-1
C 0.7 ms-1
D 0.4 ms-1
Correct Answer
Option C
Solution

Given, resistance offered by the wall

=F=−25×10−2 N=F=-25 \times 10^{-2} \mathrm{~N}

So, deacceleration of bullet,

a=Fm=−2.5×10−220×10−3=−54 ms−2(∵m=20 g=20×10−3 kg)\begin{aligned} a=\frac{F}{m}=\frac{-2.5 \times 10^{-2}}{20 \times 10^{-3}} & =-\frac{5}{4} \mathrm{~ms}^{-2} \\\\ (\because m & \left.=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\right) \end{aligned}

Now, using the equation of motion,

v2−u2=2asv^2-u^2=2 a s

We have,

v2=1+2(−54)(20×10−2)v^2=1+2\left(-\frac{5}{4}\right)\left(20 \times 10^{-2}\right)
(∵u=1 ms−1 and s=20 cm=20×10−2 m)\left(\because u=1 \mathrm{~ms}^{-1} \text{ and } s=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\right)
⇒v2=12∴v=12≈0.7 ms−1\begin{array}{ll} \Rightarrow v^2=\frac{1}{2} \\\\ \therefore v=\frac{1}{\sqrt{2}} \approx 0.7 \mathrm{~ms}^{-1} \end{array}
Q89
A block of mass mm is connected to another block of massmass MM by a spring (massless) of spring constant k.k. The block are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force FF starts acting on the block of mass MM to pull it. Find the force of the block of mass m.m.
A MF(m+M){{MF} \over {\left( {m + M} \right)}}
B mFM{{mF} \over M}
C (M+m)Fm{{\left( {M + m} \right)F} \over m}
D mF(m+M){{mF} \over {\left( {m + M} \right)}}
Correct Answer
Option D
Solution

From free body-diagram of

mm

we get

T=maT = ma

From free body-diagram of

MM

we get

F−T=MaF-T=Ma

where

TT

is force due to spring

⇒F−ma=Ma\Rightarrow F - ma = Ma

⇒\Rightarrow

F=Ma+maF=Ma+ma

∴\therefore

a=FM+ma = {F \over {M + m}}

Now, force acting on the block of mass

mm

is

ma=m(FM+m)=mFm+M.ma = m\left( {{F \over {M + m}}} \right) = {{mF} \over {m + M}}.
Q90
A body of mass 4 kg4 \mathrm{~kg} experiences two forces F⃗1=5i^+8j^+7k^\vec{F}_1=5 \hat{i}+8 \hat{j}+7 \hat{k} and F→2=3i^−4j^−3k^\overrightarrow{\mathrm{F}}_2=3 \hat{i}-4 \hat{j}-3 \hat{k}. The acceleration acting on the body is :
A 2i^+j^+k^2 \hat{i}+\hat{j}+\hat{k}
B 4i^+2j^+2k^4 \hat{i}+2 \hat{j}+2 \hat{k}
C −2i^−j^−k^-2 \hat{i}-\hat{j}-\hat{k}
D 2i^+3j^+3k^2 \hat{i}+3 \hat{j}+3 \hat{k}
Correct Answer
Option A
Solution

To find the acceleration acting on the body, we first need to determine the resultant force acting on the body by adding the two forces F⃗1\vec{F}_1 and F⃗2\vec{F}_2 vectorially.

Then, we apply Newton's second law of motion, which states that the acceleration a⃗\vec{a} of a body is directly proportional to the total force F⃗\vec{F} acting on it and inversely proportional to the mass mm of the body :

F⃗=m⋅a⃗\vec{F} = m \cdot \vec{a}

or

a⃗=F⃗m\vec{a} = \frac{\vec{F}}{m}

Let's start by adding the forces:

F⃗1+F⃗2=(5i^+8j^+7k^)+(3i^−4j^−3k^)\vec{F}_1 + \vec{F}_2 = (5 \hat{i}+8 \hat{j}+7 \hat{k}) + (3 \hat{i}-4 \hat{j}-3 \hat{k})

Performing the addition component-wise:

F⃗total=(5+3)i^+(8−4)j^+(7−3)k^ F⃗total=8i^+4j^+4k^\vec{F}_{\text{total}} = (5 + 3)\hat{i} + (8 - 4)\hat{j} + (7 - 3)\hat{k} \ \vec{F}_{\text{total}} = 8 \hat{i} + 4 \hat{j} + 4 \hat{k}

Now, let's use the formula for acceleration with m=4 kgm = 4 \mathrm{~kg}:

a⃗=F⃗totalm=8i^+4j^+4k^4 kg\vec{a} = \frac{\vec{F}_{\text{total}}}{m} = \frac{8 \hat{i} + 4 \hat{j} + 4 \hat{k}}{4 \mathrm{~kg}}

Divide each component by the mass:

a⃗=2i^+1j^+1k^\vec{a} = 2 \hat{i} + 1 \hat{j} + 1 \hat{k}

So, the acceleration acting on the body is:

a⃗=2i^+j^+k^\vec{a} = 2 \hat{i} + \hat{j} + \hat{k}

Thus, the correct option is: Option A :

2i^+j^+k^2 \hat{i}+\hat{j}+\hat{k}
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