Free Objective Test 02 Practice Test - 11th and 12th

Question 1

A smooth uniform rod of length 'L' and mass 'M' has two identical beads of negligible size, each of mass 'm', which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with angular velocity $ω_{0}$ about an axis perpendicular to the rod and passing through the mid point of the rod (see figure). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is

ω_{0}

\frac{M ω_{0}}{M + 12 m}

\frac{M ω_{0}}{M + 2 m}

\frac{M ω_{0}}{M + 6 m}

SOLUTION

Solution : D

Since there are no external forces therefore the angular momentum of the system remains constant.
Initially when the beads are at the centre of the rod angular momentum $L_{1} = (\frac{M L^{2}}{12}) ω_{0}$ .....(i)
When beads reach the ends of the rod then angular momentum $= (m {(\frac{L}{2})}^{2} + m {(\frac{L}{2})}^{2} + \frac{M L^{2}}{12}) ω^{'}$
..(ii) Equating (i) and (ii) $\frac{M L^{2}}{12} ω_{0} = (\frac{m L^{2}}{2} + \frac{M L^{2}}{12}) ω^{'} \Rightarrow ω^{'} = \frac{M ω_{0}}{M + 6 m} .$

Question 2

Moment of inertia of uniform rod of mass 'M' and length 'L' about an axis through its centre and perpendicular to its length is given by $\frac{M L^{2}}{12}$ . Now consider one such rod pivoted at its centre, free to rotate in a vertical plane. The rod is at rest in the vertical position. A bullet of mass 'M' moving horizontally at a speed 'v' strikes and embedded in one end of the rod. The angular velocity of the rod just after the collision will be

\frac{v}{L}

\frac{2 v}{L}

\frac{3 v}{2 L}

\frac{6 v}{L}

SOLUTION

Solution : C

Initial angular momentum of the system = Angular momentum of bullet before collision $= M v (\frac{L}{2})$
.....(i) let the rod rotates with angular velocity $ω$ .
Final angular momentum of the system $= (\frac{M L^{2}}{12}) ω + M {(\frac{L}{2})}^{2} ω$ ....(ii)
By equation (i) and (ii) $M v \frac{L}{2} = (\frac{M L^{2}}{12} + \frac{M L^{2}}{4}) ω o r ω = \frac{3 v}{2 L}$

Question 3

In the above problem the angular velocity of the system after the particle sticks to it will be

A. 0.3 rad/s

B. 5.3 rad/s

C. 10.3 rad/s

D. 89.3 rad/s

SOLUTION

Solution : C

Initial angular momentum of bullet + initial angular momentum of cylinder
= Final angular momentum of (bullet + cylinder) system
$\Rightarrow m v r + I_{1} ω = (I_{1} + I_{2}) ω^{'}$
$\Rightarrow m v r + I_{1} ω = (\frac{1}{2} M r^{2} + m r^{2}) ω^{'}$
$\Rightarrow 0.5 \times 5 \times 0.2 + 0.12 = (\frac{1}{2} 2 (0.2)^{2} + (0.5) (0.2)^{2}) ω^{'}$
$∴ ω^{'} = 10.3 r a d / s e c .$

Question 4

A cord is wound round the circumference of wheel of radius 'r'. The axis of the wheel is horizontal and moment of inertia about it is 'I'. A weight 'mg' is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be

\sqrt{\frac{2 g h}{I + m r}}

\sqrt{\frac{2 m g h}{I + m r^{2}}}

\sqrt{\frac{2 m g h}{I + 2 m r^{2}}}

\sqrt{2 g h}

SOLUTION

Solution : B

According to law of conservation of energy $m g h = \frac{1}{2} (I + m r^{2}) ω^{2} \Rightarrow ω = \sqrt{\frac{2 m g h}{I + m r^{2}}} .$

Question 5

A solid sphere and a disc of same mass and radius starts rolling down a rough inclined plane, from the same height the ratio of the time taken in the two cases is

15 : 14

\sqrt{15} : \sqrt{14}

14 : 15

\sqrt{14} : \sqrt{15}

SOLUTION

Solution : D

Time of descent $t = \frac{1}{sin θ} \sqrt{\frac{2 h}{g} (1 + \frac{k^{2}}{R^{2}})} ∴ R a t i o = \sqrt{\frac{{(1 + \frac{k^{2}}{R^{2}})}_{s p h e r e}}{{(1 + \frac{k^{2}}{R^{2}})}_{d i s c}}} = \sqrt{\frac{1 + \frac{2}{5}}{1 + \frac{1}{2}}} = \sqrt{\frac{7}{5} \times \frac{2}{3}} = \sqrt{\frac{14}{15}}$

Question 6

A solid sphere of mass 0.1 kg and radius 2 cm rolls down an inclined plane 1.4m in length at an angle whose slope is 1/10. If sphere started from rest, its final velocity will be

A. 1.4 m/sec

B. 0.14 m/sec

C. 14 m/sec

D. 0.7 m/sec

SOLUTION

Solution : A

$v = \sqrt{\frac{2 g h}{1 + \frac{k^{2}}{R^{2}}}} = \sqrt{\frac{2 \times 9.8 \times l sin θ}{1 + \frac{2}{5}}} [A s \frac{k^{2}}{R^{2}} = \frac{2}{5}, l = \frac{h}{s i n θ} a n d sin θ = \frac{1}{10} g i v e n]$
$\Rightarrow v = \sqrt{\frac{2 \times 9.8 \times 1.4 \times \frac{1}{10}}{\frac{7}{5}}} = 1.4 m / s .$

Question 7

A solid sphere rolls down an inclined plane and its velocity at the bottom is $v_{1}$ . Then same sphere slides down the plane (without friction) and let its velocity at the bottom be $v_{2}$ . Which of the following relation is correct

v_{1} = v_{2}

v_{1} = \frac{5}{7} v_{2}

v_{1} = \frac{7}{5} v_{2}

D. None of these

SOLUTION

Solution : D

When solid sphere rolls down an inclined plane the velocity at bottom $v_{1} = \sqrt{\frac{10}{7} g h}$
but, if there is no friction then it slides on inclined plane and the velocity at bottom $v_{2} = \sqrt{2 g h}$
$∴ \frac{v_{1}}{v_{2}} = \sqrt{\frac{5}{7}} .$

Question 8

In the following figure, a body of mass 'm' is tied at one end of a light string and this string is wrapped around the solid cylinder of mass 'M' and radius 'R'. At the moment, t = 0 the system starts moving. If the friction is negligible, angular velocity at time, t would be

\frac{m g R t}{(M + m)}

\frac{2 M g t}{R (M + 2 m)}

\frac{2 m g t}{R (M - 2 m)}

\frac{2 m g t}{R (M + 2 m)}

SOLUTION

Solution : D

We know the tangential acceleration $a = \frac{g}{1 + \frac{I}{m R^{2}}} = \frac{g}{1 + \frac{1 / 2 M R^{2}}{m R^{2}}} = \frac{2 m g}{2 m + M}$ $[A s I = \frac{1}{2} M R^{2} f o r c y l i n d e r]$
After time t, linear velocity of mass $m, v = u + a t = 0 + \frac{2 m g t}{2 m + M}$
So angular velocity of the cylinder $ω = \frac{v}{R} = \frac{2 m g t}{R (M + 2 m)} .$

Question 9

A smooth sphere A is moving on a frictionless horizontal plane with angular speed $ω$ and centre of mass is moving translationally with velocity 'v'. It collides elastically and head-on with an identical sphere B which is at rest. All surfaces are frictionless. After the collision, their angular speeds are $ω_{A}$ and $ω_{B}$ , respectively. Then,

ω_{B} = 0

ω_{A} = ω_{B}

ω_{A} < ω_{B}

ω_{B} = ω

SOLUTION

Solution : A

$ω_{B} = 0$ rotational KE can’t be transferred from A to B as surfaces are frictionless.

Question 10

A cubical block of side 'a' is moving with a velocity 'v' on a smooth horizontal plane as shown in the figure. It hits a ridge at point O. The angular speed of the block after it hits O is:

\frac{3 v}{4 a}

\frac{3 v}{2 a}

\frac{\sqrt{3} v}{\sqrt{2} a}

D. Zero

SOLUTION

Solution : A

Conserving Angular momentum before and after collision
$m v \times \frac{a}{2} = I ω$
Now, $I = \frac{m (\sqrt{2} a)^{2}}{12} + m {(\frac{a}{\sqrt{2}})}^{2}$
$= m a^{2} (\frac{1}{6} + \frac{1}{2})$
$= \frac{2}{3} m a^{2} .$
$∴ ω = \frac{m v a}{2} \times \frac{3}{2 m a^{2}} = \frac{3 v}{4 a} .$

Question 11

The torque $⃗ τ$ on a body about a given point is found to be $⃗ A \times ⃗ L$ where $⃗ A$ is a constant vector and $⃗ L$ is angular momentum of the body about that point. From this it follows that

\frac{d ⃗ L}{d t}

is perpendicular to

⃗ L

at all times.

the component of $⃗ L$ in the direction of $⃗ A$ does change with time.

the magnitude of $⃗ L$ does change with time.

⃗ L

does not change with time.

SOLUTION

Solution : A

Due to law of conservation of angular momentum, $⃗ L$ = constant
i.e. $⃗ L . ⃗ L$ = constant
or, $\frac{d}{d t} (⃗ L . ⃗ L) = 0$
or, $2 ⃗ L . \frac{d ⃗ L}{d t} = 0$
or, $⃗ L ⊥ \frac{d ⃗ L}{d t}$
Since $τ = ⃗ A \times ⃗ L$
$\frac{d ⃗ L}{d t} = ⃗ A \times ⃗ L$
i.e., $\frac{d ⃗ L}{d t}$ must be perpendicular to $⃗ A$ as well as $⃗ L$ .
Further the component of $⃗ L$ along $⃗ A$ is $\frac{⃗ A . ⃗ L}{A} .$ Also
$\frac{d}{d t} (⃗ A . ⃗ L) = ⃗ A . \frac{d ⃗ L}{d t} + ⃗ L . \frac{d ⃗ A}{d t} = 0$ ${∵ ⃗ A ⊥ \frac{d ⃗ L}{d t} a n d \frac{d ⃗ A}{d t} = ⃗ 0}$
or, $⃗ A . ⃗ L$ = constant
i.e., $\frac{⃗ A . ⃗ L}{A}$ = x = constant
Since $\frac{d ⃗ L}{d t} (o r τ)$ is perpendicular to $⃗ L$ , hence it cannot change magnitude of $⃗ L$ but can surely change direction of $⃗ L$ .

Question 12

A particle is projected at time t =0 from a point P with a speed $v_{0}$ at an angle of 45 $^{\circ}$ to the horizontal. Find the magnitude and the direction of the angular momentum of the particle about the point P at time t = $\frac{v_{0}}{g}$ .

$\frac{m v_{0}^{3}}{2 \sqrt{2} g}^j$

$\frac{m v_{0}^{3}}{2 \sqrt{2} g}^(- j)$

$\frac{m v_{0}^{3}}{\sqrt{2} g}^j$

$\frac{2 m v_{0}^{3}}{3}^(- j)$

SOLUTION

Solution : B

Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upwards direction as shown in figure. For the horizontal motion during the time 0 to t,

       $v_{x} = v_{0} c o s 45^{\circ} = \frac{v_{0}}{\sqrt{2}}$

and $x = v_{x} t = \frac{v_{0}}{\sqrt{2}} . \frac{v_{0}}{g} = \frac{v_{0}^{2}}{\sqrt{2} g}$

For vertical motion,

     $v_{γ} = v_{0} s i n 45^{\circ} - g t = \frac{v_{0}}{\sqrt{2}} - v_{0} = \frac{1 - \sqrt{2}}{\sqrt{2}} v_{0}$

and              $y = (v_{0} s i n 45^{\circ}) t - \frac{1}{2} g t^{2}$

$y = \frac{v_{0}^{2}}{\sqrt{2} g} - \frac{v_{0}^{2}}{2 g} = \frac{v_{0}^{2}}{2 g} (\sqrt{2} - 1)$

The angular momentum of the particle at time t about the origin is

                                   $L = ⃗ r \times ⃗ p = m ⃗ r \times ⃗ v$

= $m (\to i x + \to j y) \times (- \to i v_{x} + - \to j v_{y})$

                                                      = $m (⃗ k x v_{y} -_{y} v_{x})$

                                  $= m ⃗ k [(\frac{v_{0}^{2}}{\sqrt{2} g}) \frac{v_{0}}{\sqrt{2}} (1 - \sqrt{2}) - \frac{v_{0}^{2}}{2 g} (\sqrt{2} - 1) \frac{v_{0}}{\sqrt{2}}]$

                                                          $= - ⃗ k \frac{m v_{0}^{3}}{2 \sqrt{2} g}$

Thus, the angular momentum of the particle is $\frac{m v_{0}^{3}}{2 \sqrt{2} g}$ in the negative Z-direction, i.e., perpendicular to the plane of motion, going into the plane.

Question 13

A uniform rod of mass 'm' and length 'l' is kept vertical with the lower end clamped. It is slightly pushed to let it fall down under gravity. Find its angular speed when the rod is passing through its lowest position. Neglect any friction at the clamp. What will be the linear speed of the free end at this instant?

$\sqrt{\frac{6 g}{l}}, \sqrt{6 g l}$

$\sqrt{\frac{2 g}{l}}, \sqrt{6 g l}$

$\sqrt{\frac{6 g}{l}}, \sqrt{3 g l}$

$\sqrt{\frac{3 g}{l}}, \sqrt{6 g l}$

SOLUTION

Solution : A

As the rod reaches its lowest position, the centre of mass is lowered by a distance l. Its gravitational potential energy is decreased by mgl. As no energy is lost against friction, this should be equal to the increase in the kinetic energy. As the rotation occurs about the horizontal axis through the clamped end, the moment of inertia is I = $m l^{2} / 3$ . Thus,

                                    $\frac{1}{2} I ω^{2} = m g l$

                                      $\frac{1}{2} (\frac{m l^{2}}{3}) ω^{2} = m g l$

or                               $ω = \sqrt{\frac{6 g}{l}}$ .

The linear speed of the free end is

$V = I ω = \sqrt{6 g l}$ .

Question 14

A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60 $^{\circ}$ and then released. Find the magnitude of the force acting on a particle of mass 'dm' at the tip of the rod, when rod makes an angle of 37 $^{\circ}$ with the vertical.

0.9(dm)g

$0.9 \sqrt{3} (d m) g$

$0.9 \sqrt{2} (d m) g$

$(\frac{9}{10}) (\sqrt{5}) (d m) g$

SOLUTION

Solution : C

Let I = length of the rod, and m = mass of the rod.

Applying energy principle

$(\frac{1}{2}) l ω^{2} - O = m g (\frac{1}{2}) (c o s 37^{\circ} - c o s 60^{\circ})$

$\Rightarrow \frac{1}{2} \times \frac{m l^{2}}{3} ω^{2} = m g \times \frac{1}{2} (\frac{4}{5} - \frac{1}{2})$

$\Rightarrow ω^{2} = \frac{9 g}{10 l} = 0.9 (\frac{g}{l})$

Again $(\frac{\frac{m}{2}}{3}) \propto= m g (\frac{1}{2}) s i n 37^{\circ} = \frac{m g l}{2} \times \frac{3}{5}$

$∴ α = 0.9 (\frac{g}{l}) =$ angular acceleration.

So, to find out the force on the particle at the tip of the rod

$F_{1} -$ centrifugal force=(dm) $ω 2 l = 0.9 (d m) g$

$F_{t} -$ tangential force=(dm) $ω l = 0.9 (d m) g$

So, total force $F = \sqrt{(F_{1}^{2} + F_{t}^{2})} = 0.9 \sqrt{2} (d m) g$

Question 15

A cylinder is released from rest from the top of an inclined plane of inclination $θ$ and length 'l'. If the cylinder rolls without slipping, what will be its speed when it reaches the bottom?

$\sqrt{\frac{3}{4} g l c o s θ}$

$\sqrt{\frac{3}{4} g l s i n θ}$

$\sqrt{\frac{3}{4} g l t a n θ}$

$\sqrt{\frac{3}{4} g l c o t θ}$

SOLUTION

Solution : B

Let the mass of the cylinder be m and its radius r. Suppose the linear speed of the cylinder when it reaches the bottom is v. As the cylinder rolls without slipping, its angular speed about its axis is $ω = \frac{v}{r}$ . The kinetic energy at the bottom will be= $\frac{3}{4} m v^{2}$
Note: Try reducing this result on your own for practice.

Thus, $\frac{3}{4} m v^{2} = m g l s i n θ$

or, v= $\sqrt{\frac{4}{3} g l s i n θ}$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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