Practice Start Test

Laws of Motion

JEE Physics · 73 questions · Page 1 of 8 · Click an option or "Show Solution" to reveal answer

Q1
A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of unstretched lengths l1 and l2 where, l1 = nl2 and n is an integer. The ratio k1/k2 of the corresponding force constant, k1 and k2 will be :
A 1n2{1 \over {{n^2}}}
B 1n{1 \over n}
C n2
D n
Correct Answer
Option B
Solution

For a spring, k×\times

ll

= constant. ∴\therefore k1

l1{l_1}

= k2

l2{l_2}

⇒\Rightarrow

k1k2=l2l1{{{k_1}} \over {{k_2}}} = {{{l_2}} \over {{l_1}}}

=

l2nl2{{{l_2}} \over {n{l_2}}}

=

1n{1 \over n}
Q2
A block of mass M placed inside a box descends vertically with acceleration 'a'. The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of 'a' will be
A g4{g \over 4}
B g2{g \over 2}
C 3g4{3g \over 4}
D g
Correct Answer
Option C
Solution

Using Newton's second law

mg−mg4=mamg - {{mg} \over 4} = ma
⇒a=3g4\Rightarrow a = {{3g} \over 4}
Q3
A 2 kg2 \mathrm{~kg} brick begins to slide over a surface which is inclined at an angle of 45∘45^{\circ} with respect to horizontal axis. The co-efficient of static friction between their surfaces is:
A 1
B 1.7
C 0.5
D 13\dfrac{1}{\sqrt{3}}
Correct Answer
Option A
Solution
mgsin⁡θ=μ⋅mgcos⁡θ⇒tan⁡θ=μ=1\begin{aligned} & m g \sin \theta=\mu \cdot m g \cos \theta \\ & \Rightarrow \tan \theta=\mu=1 \end{aligned}
Q4
A particle moves in xx-yy plane under the influence of a force F⃗\vec{F} such that its linear momentum is p→(t)=i^cos⁡(kt)−j^sin⁡(kt)\overrightarrow{\mathrm{p}}(\mathrm{t})=\hat{i} \cos (\mathrm{kt})-\hat{j} \sin (\mathrm{kt}). If k\mathrm{k} is constant, the angle between F→\overrightarrow{\mathrm{F}} and p→\overrightarrow{\mathrm{p}} will be :
A π2\dfrac{\pi}{2}
B π3\dfrac{\pi}{3}
C π4\dfrac{\pi}{4}
D π6\dfrac{\pi}{6}
Correct Answer
Option A
Solution

To find the angle between

F⃗\vec{F}

and

p→\overrightarrow{\mathrm{p}}

, we first need to understand the relationship between force and momentum. The force

F⃗\vec{F}

acting on a particle is related to the rate of change of its linear momentum

p→\overrightarrow{\mathrm{p}}

with respect to time, as described by Newton's second law of motion:

F⃗=dp→dt\vec{F} = \frac{d\overrightarrow{\mathrm{p}}}{dt}

Given the expression for the momentum

p→(t)=i^cos⁡(kt)−j^sin⁡(kt)\overrightarrow{\mathrm{p}}(t) = \hat{i} \cos (kt) - \hat{j} \sin (kt)

, we can find

F⃗\vec{F}

by differentiating

p→\overrightarrow{\mathrm{p}}

with respect to

tt

:

dp→dt=−ki^sin⁡(kt)−kj^cos⁡(kt)\frac{d\overrightarrow{\mathrm{p}}}{dt} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)

So,

F⃗=−ki^sin⁡(kt)−kj^cos⁡(kt)\vec{F} = -k\hat{i} \sin (kt) - k\hat{j} \cos (kt)

. Now, to find the angle between

F⃗\vec{F}

and

p→\overrightarrow{\mathrm{p}}

, we use the dot product formula:

F⃗⋅p→=∣F⃗∣∣p→∣cos⁡(θ)\vec{F} \cdot \overrightarrow{\mathrm{p}} = |\vec{F}| |\overrightarrow{\mathrm{p}}| \cos(\theta)

, where θ\theta is the angle between

F⃗\vec{F}

and

p→\overrightarrow{\mathrm{p}}

. However, in this case, it's more insightful to see if

F⃗\vec{F}

and

p→\overrightarrow{\mathrm{p}}

are orthogonal (at a

π2\frac{\pi}{2}

angle to each other), because the dot product of two perpendicular vectors is zero. The dot product of

F⃗\vec{F}

and

p→\overrightarrow{\mathrm{p}}

is:

(−ki^sin⁡(kt)−kj^cos⁡(kt))⋅(i^cos⁡(kt)−j^sin⁡(kt))=(-k\hat{i} \sin (kt) - k\hat{j} \cos (kt)) \cdot (\hat{i} \cos (kt) - \hat{j} \sin (kt)) =
−ksin⁡(kt)cos⁡(kt)+kcos⁡(kt)sin⁡(kt)=0- k \sin (kt) \cos (kt) + k \cos (kt) \sin (kt) = 0

The result is zero, indicating that the angle between

F⃗\vec{F}

and

p→\overrightarrow{\mathrm{p}}

is indeed

π2\frac{\pi}{2}

. Therefore, the correct option is: Option A

π2\frac{\pi}{2}

.

Q5
A force acts for 20 s on a body of mass 20 kg, starting from rest, after which the force ceases and then body describes 50 m in the next 10 s. The value of force will be:
A 40 N
B 20 N
C 5 N
D 10 N
Correct Answer
Option C
Solution
m=20m = 20

kg

t=20t = 20

sec. Acceleration

=F20= {F \over {20}}

m/s

2^2

∴\therefore

v=u+atv = u + at
v=0+(F20)(20)v = 0 + \left( {{F \over {20}}} \right)(20)
=F= F

ms

−1^{-1}

Now for next 10 sec.

S=utS=ut
50=F(10)50=F(10)
F=5F=5
Q6
A body of mass 10 kg10 \mathrm{~kg} is moving with an initial speed of 20 m/s20 \mathrm{~m} / \mathrm{s}. The body stops after 5 s5 \mathrm{~s} due to friction between body and the floor. The value of the coefficient of friction is: (Take acceleration due to gravity g=10 ms−2g=10 \mathrm{~ms}^{-2} )
A 0.3
B 0.2
C 0.5
D 0.4
Correct Answer
Option D
Solution

a=−μga=-\mu g

∵v=u+at0=20+(−μ×10)×550μ=20μ=25=0.4\begin{aligned} & \because v=u+a t \\\\ & 0=20+(-\mu \times 10) \times 5 \\\\ & 50 \mu=20 \\\\ & \mu=\frac{2}{5}=0.4 \end{aligned}
Q7
A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point. the rope deviated at an angle of 45o at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms−-2
A 200 N
B 140 N
C 70 N
D 100 N
Correct Answer
Option D
Solution

tan 45o =

Fmg{F \over {mg}}

∴\therefore F = mg = 10 ×\times 10 = 100 N

Q8
A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid point of the rope such that the top half of the rope makes an angle of 45o with the vertical. Then F equal : (Take g = 10 ms–2 and the rope to be massless)
A 100 N
B 75 N
C 90 N
D 70 N
Correct Answer
Option A
Solution

For equilibrium, T sin 45o = F ....(1) and T cos 45o = 10g ....(2) Performing (1)/(2) we get F = 10g = 100 N

Q9
When forces F1,  F2,  F3{F_1},\,\,{F_2},\,\,{F_3} are acting on a particle of mass mm such that F2{F_2} and F3{F_3} are mutually perpendicular, then the particle remains stationary. If the force F1{F_1} is now removed then the acceleration of the particle is
A F1/m{F_1}/m
B F2F3/mF1{F_2}{F_3}/m{F_1}
C (F2−F3)/m\left( {F{}_2 - {F_3}} \right)/m
D F2/m{F_2}/m
Correct Answer
Option A
Solution

When

F1,F2{F_1},{F_2}

and

F3{F_3}

are acting on a particle then the particle remains stationary. This means that the resultant of

F1,F2{F_1},{F_2}

and

F3{F_3}

is zero. When

F1{F_1}

is removed then particle will start moving due to the force

F2{F_2}

and

F3{F_3}

in the resultant of

F2{F_2}

and

F3{F_3}

and it should be equal and opposite to

F1.{F_1}.
i.e.i.e.
∣F→2+F→3∣=∣F→1∣\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right| = \left| {{{\overrightarrow F }_1}} \right|

∴\therefore

     a=∣F→2+F→3∣m⇒a=F1m\,\,\,\,\,a = {{\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right|} \over m} \Rightarrow a = {{{F_1}} \over m}
Q10
A light string passing over a smooth light pulley connects two blocks of masses m1{m_1} and m2{m_2} (vertically). If the acceleration of the system is g/8g/8, then the ratio of the masses is
A 8:18:1
B 9:79:7
C 4:34:3
D 5:35:3
Correct Answer
Option B
Solution

Assume that, mass m1 is greater than mass m2, so the heavier mass m1 is accelerating downward and the lighter mass m2 is accelerating upwards.

For mass

m1{m_1}

the equation will be

m1{m_1}
g−T=g-T=
m1{m_1}
aa

For mass

m2{m_2}

the equation will be

T−T-
m2{m_2}
g=g=
m2{m_2}
aa

Adding those equations we get

a=(m1−m2)gm1+m2a = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}}

∴\therefore

g8=(m1−m2)gm1+m2{g \over 8} = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}}
⇒18=m1m2−1m1m2+1\Rightarrow {1 \over 8} = {{{{{m_1}} \over {{m_2}}} - 1} \over {{{{m_1}} \over {{m_2}}} + 1}}

⇒\Rightarrow

m1m2+1={{{m_1}} \over {{m_2}}} + 1 =
8(m1m2−1){8\left( {{{{m_1}} \over {{m_2}}} - 1} \right)}

⇒\Rightarrow

m1m2=97{{{m_1}} \over {{m_2}}} = {9 \over 7}
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