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Laws of Motion

JEE Physics · 73 questions · Page 2 of 8 · Click an option or "Show Solution" to reveal answer

Q11
Two forces are such that the sum of their magnitudes is 1818 NN and their resultant is 1212 NN which is perpendicular to the smaller force. Then the magnitudes of the forces are
A 12N,12N, 6N6N
B 13N,13N, 5N5N
C 10N,10N, 8N8N
D 16N16N, 2N.2N.
Correct Answer
Option B
Solution

Let the two forces be

F1{F_1}

and

F2{F_2}

and let

F2{F_2}

is smaller than

F1{F_1}

and assume

RR

is the resultant force. Given

F1+F2=18{F_1} + {F_2} = 18
      \,\,\,\,\,\,

....

(i)(i)

From the right angle triangle,

F22+R2=F12F_2^2 + {R^2} = F_1^2

or

F12−F22=R2F_1^2 - F_2^2 = {R^2}

or

(F1+F2)\left( {{F_1} + {F_2}} \right)
(F1−F2)\left( {{F_1} - {F_2}} \right)

=

R2{R^2}

or

(18)(F1−F2)\left( {18} \right)\left( {{F_1} - {F_2}} \right)

=

(12)2{\left( {12} \right)^2}

= 144 or

(F1−F2)=8\left( {{F_1} - {F_2}} \right) = 8
      \,\,\,\,\,\,

....

(ii)(ii)

By solving equation

(i)(i)

and

(ii)(ii)

we get,

F1=13 N{{F_1} = 13\,N}

and

F2=5 N{{F_2} = 5\,N}
Q12
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 4949 N,N, when the lift is stationary. If the lift moves downward with an acceleration of 5m/s25 m/{s^2}, the reading of the spring balance will be
A 2424 NN
B 7474 NN
C 1515 NN
D 4949 NN
Correct Answer
Option A
Solution

When lift is stationary then, T1 =

mgmg

= 49 N

mm

= 5 For the bag accelerating down

mg−T=mamg-T=ma

∴\therefore

T=m(g−a)T=m(g-a)
=5(9.8−5)=24 N= 5\left( {9.8 - 5} \right) = 24\,N
Q13
A rocket with a lift-off mass 3.5×104  kg3.5 \times {10^4}\,\,kg is blasted upwards with an initial acceleration of 10m/s2.10m/{s^2}. Then the initial thrust of the blast is
A 3.5×105N3.5 \times {10^5}N
B 7.0×105N7.0 \times {10^5}N
C 14.0×105N14.0 \times {10^5}N
D 1.75×105N1.75 \times {10^5}N
Correct Answer
Option A
Solution

Here, thrust force is responsible to accelerate the rocket, So initial thrust of the blast = (Lift-off mass) × acceleration = (3.5 × 104) × (10) = 3.5 × 105 N

Q14
A light spring balance hangs from the hook of the other light spring balance and a block of mass MM kgkg hangs from the former one. Then the true statement about the scale reading is
A Both the scales read MM kgkg each
B The scale of the lower one reads MM kgkg and of the upper one zero
C The reading of the two scales can be anything but the sum of the reading will be MM kgkg
D Both the scales read M/2M/2 kgkg each
Correct Answer
Option A
Solution

Question says, both spring balance are light that means springs are massless. The Earth pulls the block by a force

Mg.Mg.

The block in turn exerts a force

MgMg

on the spring of spring balance

S1{S_1}

which therefore shows a reading of

Mkgf.Mkgf.

The spring

S1{S_1}

is massless. Therefore it exerts a force of

MgMg

on the spring of spring balance

S2{S_2}

which shows the reading of

Mkgf.Mkgf.
Q15
A block rests on a rough inclined plane `making an angle of 30∘{30^ \circ } with the horizontal. The coefficient of static friction between the block and the plane is 0.8.0.8. If the frictionless force on the block is 1010 N,N, the mass of the block (in kgkg) is (take   g = 10  m/s2)\left( {take\,\,\,g\, = \,10\,\,m/{s^2}} \right)
A 1.61.6
B 4.04.0
C 2.02.0
D 2.52.5
Correct Answer
Option C
Solution

For equilibrum of block,

mg  sin⁡θ=fs  mg\,\,\sin \theta = {f_s}\,\,
⇒m×10×sin⁡30∘=10\Rightarrow m \times 10 \times \sin {30^ \circ } = 10
⇒m×5=10\Rightarrow m \times 5 = 10
⇒m=2.0  kg\Rightarrow m = 2.0\,\,kg
Q16
A particle of mass 0.3 kg subjected to a force F=−kxF=-kx with k=15k=15 N/mN/m. What will be its initial acceleration if it is released from a point 20 cm away from the origin?
A 15    m/s215\,\,\,\,m/{s^2}
B 3   m/s23\,\,\,m/{s^2}
C 10   m/s210\,\,\,m/{s^2}
D 5   m/s25\,\,\,m/{s^2}
Correct Answer
Option C
Solution

Given F = - kx ⇒\Rightarrow F = - 15

×20100\times {{20} \over {100}}

= - 3 N F = m.a = 3 N ⇒\Rightarrow a =

3m{3 \over m}

=

30.3{3 \over {0.3}}

= 10 m/s2

Q17
Consider a car moving on a straight road with a speed of 100100 m/sm/s. The distance at which car can be stopped is [μk=0.5]\left[ {{\mu _k} = 0.5} \right]
A 10001000 mm
B 800800 mm
C 400400 mm
D 100100 mm
Correct Answer
Option A
Solution

Acceleration due to friction =

(−μkg)\left( { - {\mu _k}g} \right)

We know,

v2=u2+2as{v^2} = {u^2} + 2as

⇒\Rightarrow

02=u2+2(−μkg)s{0^2} = {u^2} + 2\left( { - {\mu _k}g} \right)s

⇒\Rightarrow

2μkgs2 { {\mu _k}g}s

=

u2{u^2}
⇒s=10022×0.5×10\Rightarrow s = {{{{100}^2}} \over {2 \times 0.5 \times 10}}
⇒s=1000 m\Rightarrow s = 1000\,m
Q18
A body of mass 2 kg slides down with an acceleration of 3 m/s2 on a rough inclined plane having a slope of 30o{30^o}. The external force required to take the same body up the plane with the same acceleration will be : (g = 10 m/s2)
A 14 N
B 20 N
C 6 N
D 4 N
Correct Answer
Option B
Solution

When mass slide down then, Mgsinθ\theta −- μ\muMg cosθ\theta = Ma ⇒\Rightarrow

   \,\,\,

a = g(sin30o −- μ\mu cos30o) When mass pushed upward with force F, F −- Mgsinθ\theta −- μ\mu Mgcosθ\theta = Ma ⇒\Rightarrow

   \,\,\,

F = Mg(sin30o + μ\mu cos 30o) + Mg(sin30o −- μ\mu cos30o) ⇒\Rightarrow

   \,\,\,

F = 2Mg sin30o = 2 ×\times 2 ×\times 10 ×\times

12{1 \over 2}

= 20 N

Q19
A given object takes n times more time to slide down a 45∘{45^ \circ } rough inclined plane as it takes to slide down a perfectly smooth 45∘{45^ \circ } incline. The coefficient of kinetic friction between the object and the incline is :
A 12−n2{1 \over {2 - {n^2}}}
B 1−1n21 - {1 \over {{n^2}}}
C 1−1n2\sqrt {1 - {1 \over {{n^2}}}}
D 11−n2\sqrt {{1 \over {1 - {n^2}}}}
Correct Answer
Option B
Solution

Let, t1 and t2 are time taken to move on the smooth and rough surface for smooth surface, S =

12{1 \over 2}

g sin45o

t12t_1^2

⇒\Rightarrow

    \,\,\,\,

t1 =

22Sg\sqrt {{{2\sqrt 2 S} \over g}}

For rough surface, S =

12{1 \over 2}

g (sin45o −- μ\muk cos45o)

t22t_2^2

⇒\Rightarrow

    \,\,\,\,

t2 =

22Sg(1−μk)\sqrt {{{2\sqrt 2 S} \over {g\left( {1 - {\mu _k}} \right)}}}

Here

μk{\mu _k}

= Kinetic friction. According to question, t2 = n t1 ⇒\Rightarrow

    \,\,\,\,
22 Sg(1−μk){{2\sqrt 2 \,S} \over {g\left( {1 - {\mu _k}} \right)}}

= n2 ×\times

22 Sg{{2\sqrt 2 \,S} \over g}

⇒\Rightarrow

    \,\,\,\,

1 −- μ\muk =

1n2{1 \over {{n^2}}}

⇒\Rightarrow

    \,\,\,\,
μk{\mu _k}

= 1 −-

1n2{1 \over {{n^2}}}
Q20
A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the ball is affected by a drag force equal to mγ\gamma u2 (where m is mass of the ball, u is its instantaneous velocity and γ\gamma is a constant). Time taken by the ball to rise to its zenith is :
A 1γgtan⁡−1(γgV0){1 \over {\sqrt {\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {{\gamma \over g}} {V_0}} \right)
B 1γgln(1+γgV0){1 \over {\sqrt {\gamma g} }}{ln}\left( 1+ {\sqrt {{\gamma \over g}} {V_0}} \right)
C 1γgsin⁡−1(γgV0){1 \over {\sqrt {\gamma g} }}{\sin ^{ - 1}}\left( {\sqrt {{\gamma \over g}} {V_0}} \right)
D 12γgtan⁡−1(2γgV0){1 \over {\sqrt {2\gamma g} }}{\tan ^{ - 1}}\left( {\sqrt {{2\gamma \over g}} {V_0}} \right)
Correct Answer
Option A
Solution

Given, drag force, F=mγv2F=m \gamma v^2 ......(i) As we know, general equation of force

=ma=m a

.........(ii) Comparing Eqs. (i) and (ii), we get

a=γv2a=\gamma v^2

The net retardation of the ball when thrown vertically upward is therefore anet=−(g+γv2)=dvdta_{\text{net}} = - (g + \gamma v^2) = \dfrac{dv}{dt}, where gg is the acceleration due to gravity.

Rearranging terms gives us :

dvg+γv2=−dt\frac{dv}{g + \gamma v^2} = - dt

We now need to integrate both sides of this equation.

When the ball is thrown upward with velocity v0v_0 and reaches its zenith ( "zenith" refers to the highest point that the ball reaches in its upward trajectory.), the velocity is 00.

The time to reach the zenith is tt.

So the integral equation is :

∫v00dvγv2+g=−∫0tdt\int\limits_{v_0}^{0} \frac{dv}{\gamma v^2 + g} = - \int\limits_{0}^{t} dt

Separating the constants from the integral :

1γ∫v001(gγ+v2)dv=−∫0tdt\frac{1}{\gamma} \int\limits_{v_0}^{0} \frac{1}{\left(\frac{g}{\gamma}+v^2\right)} dv = - \int\limits_{0}^{t} dt

Recognizing the integral as the standard form 1x2+a2=1atan⁡−1(xa)\dfrac{1}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1}\left(\dfrac{x}{a}\right), we write the integral in terms of the arctangent function.

This gives us :

1γ(1gγ)[tan⁡−1(vgγ)]v00=−t\frac{1}{\gamma} \left(\frac{1}{\sqrt{\frac{g}{\gamma}}}\right) \left[\tan^{-1}\left(\frac{v}{\sqrt{\frac{g}{\gamma}}}\right)\right]_{v_0}^{0} = -t

Evaluating the integral at the bounds gives us :

1γgtan⁡−1(γv0g)=t\frac{1}{\sqrt{\gamma g}} \tan^{-1}\left(\frac{\sqrt{\gamma} v_0}{\sqrt{g}}\right) = t

Therefore, the time taken by the ball to rise to its zenith, considering the drag force, is given by

t=1γgtan⁡−1(γgV0)t = \frac{1}{\sqrt{\gamma g}} \tan^{-1}\left(\sqrt{\frac{\gamma}{g}} V_0\right)
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