QBManager

Rotational Motion

149773 A thin circular ring of mass $M$ and radius $R$ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $ω$ . Four small spheres each of mass $m$ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

1

4 ω

2

\frac{M}{4 m} ω

3

(\frac{M + 4 m}{M}) ω

4

(\frac{M}{M - 4 m}) ω

5

(\frac{M}{M + 4 m}) ω

Explanation:

E As the mass are putted gently, so no external torque is applied on the system
$∴ L = I ω =$ constant
Now,
$L_{1} = {MR}^{2} ω$ and $L_{2} = (I_{1} + I_{2}) ω^{'} [\begin{array}{l} I_{1} = {MR}^{2} \\ I_{2} = 4 {mR}^{2} \end{array}]$
$L_{2} = ({MR}^{2} + 4 {mR}^{2}) ω^{'}$
Since, angular momentum is conserved,
$L_{1} = L_{2}$
${MR}^{2} ω = (M + 4 m) R^{2} ω^{'}$
$ω^{'} = (\frac{M}{M + 4 m}) ω$

JEE Main - 26/06/2022 Shift-I]

Rotational Motion

149775 A uniform rod of length $60 cm$ is placed with one end in contact with the horizontal table and is then inclined at an angle $30^{\circ}$ to the horizontal and allowed to fall. The angular velocity of the rod when it becomes horizontal is-
$($ acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

9 rad s^{- 1}

2

6 rad s^{- 1}

3

5 rad s^{- 1}

4

8 {rad s}^{- 1}

Explanation:

C Given,
Length, $L = 60 cm = 0.6$ metre,
Angle, $θ = 30^{\circ}$
P.E. of $rod = mg$ . $\frac{L}{2} \sin 30^{\circ}$
P.E. of $rod = mg \frac{L}{2} \times \frac{1}{2}$
Rotational K.E, $\frac{I ω^{2}}{2} = \frac{1}{2} \frac{{mL}^{2}}{3} ω^{2}$
Apply law of conservation of energy,
$(Potential energy)_{initial} = (K . E)_{final}$

$\Rightarrow \frac{mgL}{2} \times \frac{1}{2} = \frac{1}{2} \times \frac{1}{3} {mL}^{2} ω^{2}$
$ω^{2} = \frac{3 g}{2 L}$
$ω = \sqrt{\frac{3 g}{2 L}} = \sqrt{\frac{3 \times 10}{2 \times 0.6}}$
$ω = \sqrt{25}$
$ω = 5 rad / \sec$

AP EAMCET-07.07.2022

Rotational Motion

149776 A fan is rotating with an angular speed 300 rpm. The fan is switched off, and it takes $80 s$ to come to rest. Assuming constant angular deceleration, the number of revolutions made by the fan before it comes to rest is

1 400

2 200

3 300

4 314

Explanation:

B Given,
Angular speed $= 300 rpm$
or, $ω_{0} = \frac{300}{60} = 5 rps$
Angular speed of fan at $t = 80 \sec$
$ω = 0 (comes to rest)$
Now, according to the question,
$ω = ω_{0} + α t$
$0 = 5 + (α \times 80)$
$- 5 = (α \times 80)$
$α = \frac{- 5}{80} rad / s^{2}$
So, the number of revolution made by the fan before is cones to rest,
$ω^{2} = ω_{0}^{2} + 2 α θ$
$0 = (5)^{2} + 2 \times (\frac{- 5}{80}) θ$
$0 = 25 - \frac{θ}{8}$
$∴ θ = 200 rev$

AP EAMCET-06.07.2022

Rotational Motion

149778 A merry - go - round rotating at a constant angular speed completes 9 rotations in 18 seconds. What is it angular speed?

1

π / 2 rad / s

2

π rad / s

3

2 π rad / s

4

3 π rad / s

Explanation:

B Given that,
$θ = 9 rev = 9 \times 2 π rad$
$t = 18 second$
We know that -
Angular speed, $ω = \frac{θ}{t}$
$ω = \frac{9 \times 2 π}{18}$
$ω = π rad / s$

TS EAMCET 19.07.2022

Rotational Motion

149779 A wheel moving with initial angular velocity 20 $rad / s$ , stops after 50 revolutions. The time taken by the wheel to stop is

1

17.1 \sec

2

15.7 \sec

3

31.4 \sec

4

47.1 \sec

Explanation:

C Given,
$ω_{0} = 20 rad / s, ω_{f} = 0, θ = 50 rev$
We know that,
$1 rev = 2 π rad$
$θ = 50 rev = 50 \times 2$
$θ = 50 \times 6.283 rad$
$θ = 314.15 rad$
$∴ θ = 50 rev = 50 \times 2 π rad$
Now,
$θ = \frac{(ω_{0} - ω_{f})}{2} \times t$
$t = \frac{2 θ}{ω_{0} - ω_{f}}$
$t = \frac{2 \times 314.15}{20 - 0}$
$t = 31.45 \sec$

TS EAMCET 30.07.2022

Rotational Motion

149773 A thin circular ring of mass $M$ and radius $R$ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $ω$ . Four small spheres each of mass $m$ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

1

4 ω

2

\frac{M}{4 m} ω

3

(\frac{M + 4 m}{M}) ω

4

(\frac{M}{M - 4 m}) ω

5

(\frac{M}{M + 4 m}) ω

Explanation:

E As the mass are putted gently, so no external torque is applied on the system
$∴ L = I ω =$ constant
Now,
$L_{1} = {MR}^{2} ω$ and $L_{2} = (I_{1} + I_{2}) ω^{'} [\begin{array}{l} I_{1} = {MR}^{2} \\ I_{2} = 4 {mR}^{2} \end{array}]$
$L_{2} = ({MR}^{2} + 4 {mR}^{2}) ω^{'}$
Since, angular momentum is conserved,
$L_{1} = L_{2}$
${MR}^{2} ω = (M + 4 m) R^{2} ω^{'}$
$ω^{'} = (\frac{M}{M + 4 m}) ω$

JEE Main - 26/06/2022 Shift-I]

Rotational Motion

149775 A uniform rod of length $60 cm$ is placed with one end in contact with the horizontal table and is then inclined at an angle $30^{\circ}$ to the horizontal and allowed to fall. The angular velocity of the rod when it becomes horizontal is-
$($ acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

9 rad s^{- 1}

2

6 rad s^{- 1}

3

5 rad s^{- 1}

4

8 {rad s}^{- 1}

Explanation:

C Given,
Length, $L = 60 cm = 0.6$ metre,
Angle, $θ = 30^{\circ}$
P.E. of $rod = mg$ . $\frac{L}{2} \sin 30^{\circ}$
P.E. of $rod = mg \frac{L}{2} \times \frac{1}{2}$
Rotational K.E, $\frac{I ω^{2}}{2} = \frac{1}{2} \frac{{mL}^{2}}{3} ω^{2}$
Apply law of conservation of energy,
$(Potential energy)_{initial} = (K . E)_{final}$

$\Rightarrow \frac{mgL}{2} \times \frac{1}{2} = \frac{1}{2} \times \frac{1}{3} {mL}^{2} ω^{2}$
$ω^{2} = \frac{3 g}{2 L}$
$ω = \sqrt{\frac{3 g}{2 L}} = \sqrt{\frac{3 \times 10}{2 \times 0.6}}$
$ω = \sqrt{25}$
$ω = 5 rad / \sec$

AP EAMCET-07.07.2022

Rotational Motion

149776 A fan is rotating with an angular speed 300 rpm. The fan is switched off, and it takes $80 s$ to come to rest. Assuming constant angular deceleration, the number of revolutions made by the fan before it comes to rest is

1 400

2 200

3 300

4 314

Explanation:

B Given,
Angular speed $= 300 rpm$
or, $ω_{0} = \frac{300}{60} = 5 rps$
Angular speed of fan at $t = 80 \sec$
$ω = 0 (comes to rest)$
Now, according to the question,
$ω = ω_{0} + α t$
$0 = 5 + (α \times 80)$
$- 5 = (α \times 80)$
$α = \frac{- 5}{80} rad / s^{2}$
So, the number of revolution made by the fan before is cones to rest,
$ω^{2} = ω_{0}^{2} + 2 α θ$
$0 = (5)^{2} + 2 \times (\frac{- 5}{80}) θ$
$0 = 25 - \frac{θ}{8}$
$∴ θ = 200 rev$

AP EAMCET-06.07.2022

Rotational Motion

149778 A merry - go - round rotating at a constant angular speed completes 9 rotations in 18 seconds. What is it angular speed?

1

π / 2 rad / s

2

π rad / s

3

2 π rad / s

4

3 π rad / s

Explanation:

B Given that,
$θ = 9 rev = 9 \times 2 π rad$
$t = 18 second$
We know that -
Angular speed, $ω = \frac{θ}{t}$
$ω = \frac{9 \times 2 π}{18}$
$ω = π rad / s$

TS EAMCET 19.07.2022

Rotational Motion

149779 A wheel moving with initial angular velocity 20 $rad / s$ , stops after 50 revolutions. The time taken by the wheel to stop is

1

17.1 \sec

2

15.7 \sec

3

31.4 \sec

4

47.1 \sec

Explanation:

C Given,
$ω_{0} = 20 rad / s, ω_{f} = 0, θ = 50 rev$
We know that,
$1 rev = 2 π rad$
$θ = 50 rev = 50 \times 2$
$θ = 50 \times 6.283 rad$
$θ = 314.15 rad$
$∴ θ = 50 rev = 50 \times 2 π rad$
Now,
$θ = \frac{(ω_{0} - ω_{f})}{2} \times t$
$t = \frac{2 θ}{ω_{0} - ω_{f}}$
$t = \frac{2 \times 314.15}{20 - 0}$
$t = 31.45 \sec$

TS EAMCET 30.07.2022

Rotational Motion

149773 A thin circular ring of mass $M$ and radius $R$ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $ω$ . Four small spheres each of mass $m$ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

1

4 ω

2

\frac{M}{4 m} ω

3

(\frac{M + 4 m}{M}) ω

4

(\frac{M}{M - 4 m}) ω

5

(\frac{M}{M + 4 m}) ω

Explanation:

E As the mass are putted gently, so no external torque is applied on the system
$∴ L = I ω =$ constant
Now,
$L_{1} = {MR}^{2} ω$ and $L_{2} = (I_{1} + I_{2}) ω^{'} [\begin{array}{l} I_{1} = {MR}^{2} \\ I_{2} = 4 {mR}^{2} \end{array}]$
$L_{2} = ({MR}^{2} + 4 {mR}^{2}) ω^{'}$
Since, angular momentum is conserved,
$L_{1} = L_{2}$
${MR}^{2} ω = (M + 4 m) R^{2} ω^{'}$
$ω^{'} = (\frac{M}{M + 4 m}) ω$

JEE Main - 26/06/2022 Shift-I]

Rotational Motion

149775 A uniform rod of length $60 cm$ is placed with one end in contact with the horizontal table and is then inclined at an angle $30^{\circ}$ to the horizontal and allowed to fall. The angular velocity of the rod when it becomes horizontal is-
$($ acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

9 rad s^{- 1}

2

6 rad s^{- 1}

3

5 rad s^{- 1}

4

8 {rad s}^{- 1}

Explanation:

C Given,
Length, $L = 60 cm = 0.6$ metre,
Angle, $θ = 30^{\circ}$
P.E. of $rod = mg$ . $\frac{L}{2} \sin 30^{\circ}$
P.E. of $rod = mg \frac{L}{2} \times \frac{1}{2}$
Rotational K.E, $\frac{I ω^{2}}{2} = \frac{1}{2} \frac{{mL}^{2}}{3} ω^{2}$
Apply law of conservation of energy,
$(Potential energy)_{initial} = (K . E)_{final}$

$\Rightarrow \frac{mgL}{2} \times \frac{1}{2} = \frac{1}{2} \times \frac{1}{3} {mL}^{2} ω^{2}$
$ω^{2} = \frac{3 g}{2 L}$
$ω = \sqrt{\frac{3 g}{2 L}} = \sqrt{\frac{3 \times 10}{2 \times 0.6}}$
$ω = \sqrt{25}$
$ω = 5 rad / \sec$

AP EAMCET-07.07.2022

Rotational Motion

149776 A fan is rotating with an angular speed 300 rpm. The fan is switched off, and it takes $80 s$ to come to rest. Assuming constant angular deceleration, the number of revolutions made by the fan before it comes to rest is

1 400

2 200

3 300

4 314

Explanation:

B Given,
Angular speed $= 300 rpm$
or, $ω_{0} = \frac{300}{60} = 5 rps$
Angular speed of fan at $t = 80 \sec$
$ω = 0 (comes to rest)$
Now, according to the question,
$ω = ω_{0} + α t$
$0 = 5 + (α \times 80)$
$- 5 = (α \times 80)$
$α = \frac{- 5}{80} rad / s^{2}$
So, the number of revolution made by the fan before is cones to rest,
$ω^{2} = ω_{0}^{2} + 2 α θ$
$0 = (5)^{2} + 2 \times (\frac{- 5}{80}) θ$
$0 = 25 - \frac{θ}{8}$
$∴ θ = 200 rev$

AP EAMCET-06.07.2022

Rotational Motion

149778 A merry - go - round rotating at a constant angular speed completes 9 rotations in 18 seconds. What is it angular speed?

1

π / 2 rad / s

2

π rad / s

3

2 π rad / s

4

3 π rad / s

Explanation:

B Given that,
$θ = 9 rev = 9 \times 2 π rad$
$t = 18 second$
We know that -
Angular speed, $ω = \frac{θ}{t}$
$ω = \frac{9 \times 2 π}{18}$
$ω = π rad / s$

TS EAMCET 19.07.2022

Rotational Motion

149779 A wheel moving with initial angular velocity 20 $rad / s$ , stops after 50 revolutions. The time taken by the wheel to stop is

1

17.1 \sec

2

15.7 \sec

3

31.4 \sec

4

47.1 \sec

Explanation:

C Given,
$ω_{0} = 20 rad / s, ω_{f} = 0, θ = 50 rev$
We know that,
$1 rev = 2 π rad$
$θ = 50 rev = 50 \times 2$
$θ = 50 \times 6.283 rad$
$θ = 314.15 rad$
$∴ θ = 50 rev = 50 \times 2 π rad$
Now,
$θ = \frac{(ω_{0} - ω_{f})}{2} \times t$
$t = \frac{2 θ}{ω_{0} - ω_{f}}$
$t = \frac{2 \times 314.15}{20 - 0}$
$t = 31.45 \sec$

TS EAMCET 30.07.2022

Rotational Motion

149773 A thin circular ring of mass $M$ and radius $R$ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $ω$ . Four small spheres each of mass $m$ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

1

4 ω

2

\frac{M}{4 m} ω

3

(\frac{M + 4 m}{M}) ω

4

(\frac{M}{M - 4 m}) ω

5

(\frac{M}{M + 4 m}) ω

Explanation:

E As the mass are putted gently, so no external torque is applied on the system
$∴ L = I ω =$ constant
Now,
$L_{1} = {MR}^{2} ω$ and $L_{2} = (I_{1} + I_{2}) ω^{'} [\begin{array}{l} I_{1} = {MR}^{2} \\ I_{2} = 4 {mR}^{2} \end{array}]$
$L_{2} = ({MR}^{2} + 4 {mR}^{2}) ω^{'}$
Since, angular momentum is conserved,
$L_{1} = L_{2}$
${MR}^{2} ω = (M + 4 m) R^{2} ω^{'}$
$ω^{'} = (\frac{M}{M + 4 m}) ω$

JEE Main - 26/06/2022 Shift-I]

Rotational Motion

149775 A uniform rod of length $60 cm$ is placed with one end in contact with the horizontal table and is then inclined at an angle $30^{\circ}$ to the horizontal and allowed to fall. The angular velocity of the rod when it becomes horizontal is-
$($ acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

9 rad s^{- 1}

2

6 rad s^{- 1}

3

5 rad s^{- 1}

4

8 {rad s}^{- 1}

Explanation:

C Given,
Length, $L = 60 cm = 0.6$ metre,
Angle, $θ = 30^{\circ}$
P.E. of $rod = mg$ . $\frac{L}{2} \sin 30^{\circ}$
P.E. of $rod = mg \frac{L}{2} \times \frac{1}{2}$
Rotational K.E, $\frac{I ω^{2}}{2} = \frac{1}{2} \frac{{mL}^{2}}{3} ω^{2}$
Apply law of conservation of energy,
$(Potential energy)_{initial} = (K . E)_{final}$

$\Rightarrow \frac{mgL}{2} \times \frac{1}{2} = \frac{1}{2} \times \frac{1}{3} {mL}^{2} ω^{2}$
$ω^{2} = \frac{3 g}{2 L}$
$ω = \sqrt{\frac{3 g}{2 L}} = \sqrt{\frac{3 \times 10}{2 \times 0.6}}$
$ω = \sqrt{25}$
$ω = 5 rad / \sec$

AP EAMCET-07.07.2022

Rotational Motion

149776 A fan is rotating with an angular speed 300 rpm. The fan is switched off, and it takes $80 s$ to come to rest. Assuming constant angular deceleration, the number of revolutions made by the fan before it comes to rest is

1 400

2 200

3 300

4 314

Explanation:

B Given,
Angular speed $= 300 rpm$
or, $ω_{0} = \frac{300}{60} = 5 rps$
Angular speed of fan at $t = 80 \sec$
$ω = 0 (comes to rest)$
Now, according to the question,
$ω = ω_{0} + α t$
$0 = 5 + (α \times 80)$
$- 5 = (α \times 80)$
$α = \frac{- 5}{80} rad / s^{2}$
So, the number of revolution made by the fan before is cones to rest,
$ω^{2} = ω_{0}^{2} + 2 α θ$
$0 = (5)^{2} + 2 \times (\frac{- 5}{80}) θ$
$0 = 25 - \frac{θ}{8}$
$∴ θ = 200 rev$

AP EAMCET-06.07.2022

Rotational Motion

149778 A merry - go - round rotating at a constant angular speed completes 9 rotations in 18 seconds. What is it angular speed?

1

π / 2 rad / s

2

π rad / s

3

2 π rad / s

4

3 π rad / s

Explanation:

B Given that,
$θ = 9 rev = 9 \times 2 π rad$
$t = 18 second$
We know that -
Angular speed, $ω = \frac{θ}{t}$
$ω = \frac{9 \times 2 π}{18}$
$ω = π rad / s$

TS EAMCET 19.07.2022

Rotational Motion

149779 A wheel moving with initial angular velocity 20 $rad / s$ , stops after 50 revolutions. The time taken by the wheel to stop is

1

17.1 \sec

2

15.7 \sec

3

31.4 \sec

4

47.1 \sec

Explanation:

C Given,
$ω_{0} = 20 rad / s, ω_{f} = 0, θ = 50 rev$
We know that,
$1 rev = 2 π rad$
$θ = 50 rev = 50 \times 2$
$θ = 50 \times 6.283 rad$
$θ = 314.15 rad$
$∴ θ = 50 rev = 50 \times 2 π rad$
Now,
$θ = \frac{(ω_{0} - ω_{f})}{2} \times t$
$t = \frac{2 θ}{ω_{0} - ω_{f}}$
$t = \frac{2 \times 314.15}{20 - 0}$
$t = 31.45 \sec$

TS EAMCET 30.07.2022

Rotational Motion

149773 A thin circular ring of mass $M$ and radius $R$ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $ω$ . Four small spheres each of mass $m$ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

1

4 ω

2

\frac{M}{4 m} ω

3

(\frac{M + 4 m}{M}) ω

4

(\frac{M}{M - 4 m}) ω

5

(\frac{M}{M + 4 m}) ω

Explanation:

E As the mass are putted gently, so no external torque is applied on the system
$∴ L = I ω =$ constant
Now,
$L_{1} = {MR}^{2} ω$ and $L_{2} = (I_{1} + I_{2}) ω^{'} [\begin{array}{l} I_{1} = {MR}^{2} \\ I_{2} = 4 {mR}^{2} \end{array}]$
$L_{2} = ({MR}^{2} + 4 {mR}^{2}) ω^{'}$
Since, angular momentum is conserved,
$L_{1} = L_{2}$
${MR}^{2} ω = (M + 4 m) R^{2} ω^{'}$
$ω^{'} = (\frac{M}{M + 4 m}) ω$

JEE Main - 26/06/2022 Shift-I]

Rotational Motion

149775 A uniform rod of length $60 cm$ is placed with one end in contact with the horizontal table and is then inclined at an angle $30^{\circ}$ to the horizontal and allowed to fall. The angular velocity of the rod when it becomes horizontal is-
$($ acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

9 rad s^{- 1}

2

6 rad s^{- 1}

3

5 rad s^{- 1}

4

8 {rad s}^{- 1}

Explanation:

C Given,
Length, $L = 60 cm = 0.6$ metre,
Angle, $θ = 30^{\circ}$
P.E. of $rod = mg$ . $\frac{L}{2} \sin 30^{\circ}$
P.E. of $rod = mg \frac{L}{2} \times \frac{1}{2}$
Rotational K.E, $\frac{I ω^{2}}{2} = \frac{1}{2} \frac{{mL}^{2}}{3} ω^{2}$
Apply law of conservation of energy,
$(Potential energy)_{initial} = (K . E)_{final}$

$\Rightarrow \frac{mgL}{2} \times \frac{1}{2} = \frac{1}{2} \times \frac{1}{3} {mL}^{2} ω^{2}$
$ω^{2} = \frac{3 g}{2 L}$
$ω = \sqrt{\frac{3 g}{2 L}} = \sqrt{\frac{3 \times 10}{2 \times 0.6}}$
$ω = \sqrt{25}$
$ω = 5 rad / \sec$

AP EAMCET-07.07.2022

Rotational Motion

149776 A fan is rotating with an angular speed 300 rpm. The fan is switched off, and it takes $80 s$ to come to rest. Assuming constant angular deceleration, the number of revolutions made by the fan before it comes to rest is

1 400

2 200

3 300

4 314

Explanation:

B Given,
Angular speed $= 300 rpm$
or, $ω_{0} = \frac{300}{60} = 5 rps$
Angular speed of fan at $t = 80 \sec$
$ω = 0 (comes to rest)$
Now, according to the question,
$ω = ω_{0} + α t$
$0 = 5 + (α \times 80)$
$- 5 = (α \times 80)$
$α = \frac{- 5}{80} rad / s^{2}$
So, the number of revolution made by the fan before is cones to rest,
$ω^{2} = ω_{0}^{2} + 2 α θ$
$0 = (5)^{2} + 2 \times (\frac{- 5}{80}) θ$
$0 = 25 - \frac{θ}{8}$
$∴ θ = 200 rev$

AP EAMCET-06.07.2022

Rotational Motion

149778 A merry - go - round rotating at a constant angular speed completes 9 rotations in 18 seconds. What is it angular speed?

1

π / 2 rad / s

2

π rad / s

3

2 π rad / s

4

3 π rad / s

Explanation:

B Given that,
$θ = 9 rev = 9 \times 2 π rad$
$t = 18 second$
We know that -
Angular speed, $ω = \frac{θ}{t}$
$ω = \frac{9 \times 2 π}{18}$
$ω = π rad / s$

TS EAMCET 19.07.2022

Rotational Motion

149779 A wheel moving with initial angular velocity 20 $rad / s$ , stops after 50 revolutions. The time taken by the wheel to stop is

1

17.1 \sec

2

15.7 \sec

3

31.4 \sec

4

47.1 \sec

Explanation:

C Given,
$ω_{0} = 20 rad / s, ω_{f} = 0, θ = 50 rev$
We know that,
$1 rev = 2 π rad$
$θ = 50 rev = 50 \times 2$
$θ = 50 \times 6.283 rad$
$θ = 314.15 rad$
$∴ θ = 50 rev = 50 \times 2 π rad$
Now,
$θ = \frac{(ω_{0} - ω_{f})}{2} \times t$
$t = \frac{2 θ}{ω_{0} - ω_{f}}$
$t = \frac{2 \times 314.15}{20 - 0}$
$t = 31.45 \sec$

TS EAMCET 30.07.2022

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