QBManager

Rotational Motion

150160 A uniform bar of mass $M$ and length $L$ is bent in the form of an equilateral triangle. Find the moment of inertia of the triangle about an axis passing through the centre of mass and perpendicular to the plane of the triangle

1

{ML}^{2}

2

{ML}^{2} / 2

3

{ML}^{2} / 27

4

{ML}^{2} / 54

Explanation:

D original image
$\tan 30^{\circ} = \frac{x}{L / 6} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{L / 6}$
$x = \frac{L}{6 \sqrt{3}}$
Apply parallel axis theorem,
$I_{Net} = I_{G} + {mx}^{2}$
$= [\frac{1}{12} \times \frac{M}{3} {(\frac{L}{3})}^{2} + \frac{M}{3} {(\frac{L}{6 \sqrt{3}})}^{2}] \times 3$
$= [\frac{M}{36} \times \frac{L^{2}}{9} + \frac{M}{3} \times \frac{L^{2}}{36 \times 3}] \times 3$
$= \frac{M L^{2}}{108} + \frac{M L^{2}}{108}$
$= \frac{2 M L^{2}}{108}$
$I_{N e t} = \frac{M L^{2}}{54}$

J and K-CET-2019

Rotational Motion

150161 A long cylindrical rod is welded to a thin circular disc of diameter $0.5 m$ at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in $m$ is

1

\frac{1}{4}

2

\frac{\sqrt{5}}{8}

3

\frac{1}{2}

4

2 \sqrt{2}

Explanation:

B Given, diameter of disc $= 0.5$ meter
$∴$ Radius of disc $= \frac{0.5}{2} = \frac{1}{4} m$

Moment of inertia of a disc about its diameter,
$I_{CG} = \frac{{MR}^{2}}{4}$
Moment of inertia of a disc about a tangent passing through circumference and parallel to its diameter.
$∴ I = \frac{{MR}^{2}}{4} + {MR}^{2} = \frac{5 {MR}^{2}}{4}$
${MK}^{2} = \frac{5}{4} {MR}^{2} [∵ I = {MK}^{2}]$
$K^{2} = \frac{5}{4} R^{2}$
$K = \sqrt{\frac{5}{4}} \times \frac{0.5}{2}$
$= \frac{\sqrt{5}}{2} \times \frac{1}{4}$
$K = \frac{\sqrt{5}}{8} m$

TS- EAMCET-03.05.2019

Rotational Motion

150162 A wheel of radius $8 cm$ is attached to a support so as to rotate about a horizontal axis through its centre. A string of negligible mass wrapped around its circumference carries a mass of 0.4 $kg$ attached to its free end. When the mass is released, its descends through $1 m$ in 10 seconds, then its moment of inertia is (Acceleration due to gravitv, $g = 10 {m s}^{- 2}$
original image

1

1.277 kg m_{2}^{2}

2

2.177 kg m^{2}

3

21.77 kg m^{2}

4

12.77 kg m^{2}

Explanation:

A Given,
Radius of wheel $(r) = 8 cm = 8 \times 10^{- 2} m$ ,
Mass of body $(m) = 0.4 kg$
Descending length $(s) = 1 m$ , Time $t) = 10 \sec$ .
original image
From second law of motion.
$s = ut + \frac{1}{2} {at}^{2}$
$1 = 0 + \frac{1}{2} a \times (10)^{2}$
$a = \frac{1}{50} m / s^{2}$
Angular acceleration,
$α = \frac{a}{r} [∵ a = α r]$
$α = \frac{\frac{1}{50}}{8 \times 10^{- 2}}$
$α = \frac{100}{50 \times 8}$
$α = \frac{1}{4} rad / s^{2}$
Velocity after $10 \sec$ ,
$v = u + at$
$v = 0 + \frac{1}{50} \times 10$
$v = 0.2 m / s$
Angular velocity,
$ω = \frac{v}{r} [∵ v = r ω]$
$ω = \frac{0.2}{8 \times 10^{- 2}} = \frac{20}{8}$
$ω = \frac{5}{2} rad / s$
From law of conservation of energy,
Loss of P.E. $=$ K.E. of translation + K.E. of rotation
$mgh = \frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2}$
$0.4 \times 10 \times 1 = \frac{1}{2} \times 0.4 \times (0.2)^{2} + \frac{1}{2} \times I \times {(\frac{5}{2})}^{2}$
$4 = (0.2 \times 0.04) + \frac{1}{2} \times I \times \frac{25}{4}$
$4 - 0.008 = 3.125 I$
$3.992 = 3.125 I$
$I = \frac{3.992}{3.125}$
$I = 1.277 kg - m^{2}$

AP EAMCET (23.04.2019) Shift-I

Rotational Motion

150163 A thin wire of length $l$ having a linear density $ρ$ is bent into a circular loop with $C$ as its centre as shown in the figure. The moment of inertia of the loop about the line $A B$ is

1

\frac{5}{16} \frac{ρ l^{3}}{π^{3}}

2

\frac{1}{16} \frac{ρ l^{3}}{π^{3}}

3

\frac{1}{8} \frac{ρ l^{3}}{π^{3}}

4

\frac{3}{8} \frac{ρ l^{3}}{π^{2}}

Explanation:

D Given,
Length of thin wire $= l$
Linear density of thin wire $= ρ$
Mass of thin wire, $M) = l ρ$
Circumference $=$ length of thin wire
$2 π R = l$
$R = \frac{l}{2 π}$
original image
The moment of inertia of a thin loop about it's diameter,
$I = \frac{1}{2} {MR}^{2}$
$I = \frac{1}{2} l ρ \times {(\frac{l}{2 π})}^{2}$
$I = \frac{l^{3} ρ}{8 π^{2}}$
Now using parallel axis theorem,
$I_{AB} = I_{C} + {MR}^{2}$
$I_{AB} = \frac{l^{3} ρ}{8 π^{2}} + l ρ \times {(\frac{l}{2 π})}^{2}$
$I_{AB} = \frac{l^{3} ρ}{8 π^{2}} + \frac{ρ l^{3}}{4 π^{2}}$
$I_{AB} = \frac{3}{8} \frac{ρ l^{3}}{π^{2}}$

AP EAMCET (22.04.2019) Shift-I

Rotational Motion

150160 A uniform bar of mass $M$ and length $L$ is bent in the form of an equilateral triangle. Find the moment of inertia of the triangle about an axis passing through the centre of mass and perpendicular to the plane of the triangle

1

{ML}^{2}

2

{ML}^{2} / 2

3

{ML}^{2} / 27

4

{ML}^{2} / 54

Explanation:

D original image
$\tan 30^{\circ} = \frac{x}{L / 6} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{L / 6}$
$x = \frac{L}{6 \sqrt{3}}$
Apply parallel axis theorem,
$I_{Net} = I_{G} + {mx}^{2}$
$= [\frac{1}{12} \times \frac{M}{3} {(\frac{L}{3})}^{2} + \frac{M}{3} {(\frac{L}{6 \sqrt{3}})}^{2}] \times 3$
$= [\frac{M}{36} \times \frac{L^{2}}{9} + \frac{M}{3} \times \frac{L^{2}}{36 \times 3}] \times 3$
$= \frac{M L^{2}}{108} + \frac{M L^{2}}{108}$
$= \frac{2 M L^{2}}{108}$
$I_{N e t} = \frac{M L^{2}}{54}$

J and K-CET-2019

Rotational Motion

150161 A long cylindrical rod is welded to a thin circular disc of diameter $0.5 m$ at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in $m$ is

1

\frac{1}{4}

2

\frac{\sqrt{5}}{8}

3

\frac{1}{2}

4

2 \sqrt{2}

Explanation:

B Given, diameter of disc $= 0.5$ meter
$∴$ Radius of disc $= \frac{0.5}{2} = \frac{1}{4} m$

Moment of inertia of a disc about its diameter,
$I_{CG} = \frac{{MR}^{2}}{4}$
Moment of inertia of a disc about a tangent passing through circumference and parallel to its diameter.
$∴ I = \frac{{MR}^{2}}{4} + {MR}^{2} = \frac{5 {MR}^{2}}{4}$
${MK}^{2} = \frac{5}{4} {MR}^{2} [∵ I = {MK}^{2}]$
$K^{2} = \frac{5}{4} R^{2}$
$K = \sqrt{\frac{5}{4}} \times \frac{0.5}{2}$
$= \frac{\sqrt{5}}{2} \times \frac{1}{4}$
$K = \frac{\sqrt{5}}{8} m$

TS- EAMCET-03.05.2019

Rotational Motion

150162 A wheel of radius $8 cm$ is attached to a support so as to rotate about a horizontal axis through its centre. A string of negligible mass wrapped around its circumference carries a mass of 0.4 $kg$ attached to its free end. When the mass is released, its descends through $1 m$ in 10 seconds, then its moment of inertia is (Acceleration due to gravitv, $g = 10 {m s}^{- 2}$
original image

1

1.277 kg m_{2}^{2}

2

2.177 kg m^{2}

3

21.77 kg m^{2}

4

12.77 kg m^{2}

Explanation:

A Given,
Radius of wheel $(r) = 8 cm = 8 \times 10^{- 2} m$ ,
Mass of body $(m) = 0.4 kg$
Descending length $(s) = 1 m$ , Time $t) = 10 \sec$ .
original image
From second law of motion.
$s = ut + \frac{1}{2} {at}^{2}$
$1 = 0 + \frac{1}{2} a \times (10)^{2}$
$a = \frac{1}{50} m / s^{2}$
Angular acceleration,
$α = \frac{a}{r} [∵ a = α r]$
$α = \frac{\frac{1}{50}}{8 \times 10^{- 2}}$
$α = \frac{100}{50 \times 8}$
$α = \frac{1}{4} rad / s^{2}$
Velocity after $10 \sec$ ,
$v = u + at$
$v = 0 + \frac{1}{50} \times 10$
$v = 0.2 m / s$
Angular velocity,
$ω = \frac{v}{r} [∵ v = r ω]$
$ω = \frac{0.2}{8 \times 10^{- 2}} = \frac{20}{8}$
$ω = \frac{5}{2} rad / s$
From law of conservation of energy,
Loss of P.E. $=$ K.E. of translation + K.E. of rotation
$mgh = \frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2}$
$0.4 \times 10 \times 1 = \frac{1}{2} \times 0.4 \times (0.2)^{2} + \frac{1}{2} \times I \times {(\frac{5}{2})}^{2}$
$4 = (0.2 \times 0.04) + \frac{1}{2} \times I \times \frac{25}{4}$
$4 - 0.008 = 3.125 I$
$3.992 = 3.125 I$
$I = \frac{3.992}{3.125}$
$I = 1.277 kg - m^{2}$

AP EAMCET (23.04.2019) Shift-I

Rotational Motion

150163 A thin wire of length $l$ having a linear density $ρ$ is bent into a circular loop with $C$ as its centre as shown in the figure. The moment of inertia of the loop about the line $A B$ is

1

\frac{5}{16} \frac{ρ l^{3}}{π^{3}}

2

\frac{1}{16} \frac{ρ l^{3}}{π^{3}}

3

\frac{1}{8} \frac{ρ l^{3}}{π^{3}}

4

\frac{3}{8} \frac{ρ l^{3}}{π^{2}}

Explanation:

D Given,
Length of thin wire $= l$
Linear density of thin wire $= ρ$
Mass of thin wire, $M) = l ρ$
Circumference $=$ length of thin wire
$2 π R = l$
$R = \frac{l}{2 π}$
original image
The moment of inertia of a thin loop about it's diameter,
$I = \frac{1}{2} {MR}^{2}$
$I = \frac{1}{2} l ρ \times {(\frac{l}{2 π})}^{2}$
$I = \frac{l^{3} ρ}{8 π^{2}}$
Now using parallel axis theorem,
$I_{AB} = I_{C} + {MR}^{2}$
$I_{AB} = \frac{l^{3} ρ}{8 π^{2}} + l ρ \times {(\frac{l}{2 π})}^{2}$
$I_{AB} = \frac{l^{3} ρ}{8 π^{2}} + \frac{ρ l^{3}}{4 π^{2}}$
$I_{AB} = \frac{3}{8} \frac{ρ l^{3}}{π^{2}}$

AP EAMCET (22.04.2019) Shift-I

Rotational Motion

150160 A uniform bar of mass $M$ and length $L$ is bent in the form of an equilateral triangle. Find the moment of inertia of the triangle about an axis passing through the centre of mass and perpendicular to the plane of the triangle

1

{ML}^{2}

2

{ML}^{2} / 2

3

{ML}^{2} / 27

4

{ML}^{2} / 54

Explanation:

D original image
$\tan 30^{\circ} = \frac{x}{L / 6} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{L / 6}$
$x = \frac{L}{6 \sqrt{3}}$
Apply parallel axis theorem,
$I_{Net} = I_{G} + {mx}^{2}$
$= [\frac{1}{12} \times \frac{M}{3} {(\frac{L}{3})}^{2} + \frac{M}{3} {(\frac{L}{6 \sqrt{3}})}^{2}] \times 3$
$= [\frac{M}{36} \times \frac{L^{2}}{9} + \frac{M}{3} \times \frac{L^{2}}{36 \times 3}] \times 3$
$= \frac{M L^{2}}{108} + \frac{M L^{2}}{108}$
$= \frac{2 M L^{2}}{108}$
$I_{N e t} = \frac{M L^{2}}{54}$

J and K-CET-2019

Rotational Motion

150161 A long cylindrical rod is welded to a thin circular disc of diameter $0.5 m$ at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in $m$ is

1

\frac{1}{4}

2

\frac{\sqrt{5}}{8}

3

\frac{1}{2}

4

2 \sqrt{2}

Explanation:

B Given, diameter of disc $= 0.5$ meter
$∴$ Radius of disc $= \frac{0.5}{2} = \frac{1}{4} m$

Moment of inertia of a disc about its diameter,
$I_{CG} = \frac{{MR}^{2}}{4}$
Moment of inertia of a disc about a tangent passing through circumference and parallel to its diameter.
$∴ I = \frac{{MR}^{2}}{4} + {MR}^{2} = \frac{5 {MR}^{2}}{4}$
${MK}^{2} = \frac{5}{4} {MR}^{2} [∵ I = {MK}^{2}]$
$K^{2} = \frac{5}{4} R^{2}$
$K = \sqrt{\frac{5}{4}} \times \frac{0.5}{2}$
$= \frac{\sqrt{5}}{2} \times \frac{1}{4}$
$K = \frac{\sqrt{5}}{8} m$

TS- EAMCET-03.05.2019

Rotational Motion

150162 A wheel of radius $8 cm$ is attached to a support so as to rotate about a horizontal axis through its centre. A string of negligible mass wrapped around its circumference carries a mass of 0.4 $kg$ attached to its free end. When the mass is released, its descends through $1 m$ in 10 seconds, then its moment of inertia is (Acceleration due to gravitv, $g = 10 {m s}^{- 2}$
original image

1

1.277 kg m_{2}^{2}

2

2.177 kg m^{2}

3

21.77 kg m^{2}

4

12.77 kg m^{2}

Explanation:

A Given,
Radius of wheel $(r) = 8 cm = 8 \times 10^{- 2} m$ ,
Mass of body $(m) = 0.4 kg$
Descending length $(s) = 1 m$ , Time $t) = 10 \sec$ .
original image
From second law of motion.
$s = ut + \frac{1}{2} {at}^{2}$
$1 = 0 + \frac{1}{2} a \times (10)^{2}$
$a = \frac{1}{50} m / s^{2}$
Angular acceleration,
$α = \frac{a}{r} [∵ a = α r]$
$α = \frac{\frac{1}{50}}{8 \times 10^{- 2}}$
$α = \frac{100}{50 \times 8}$
$α = \frac{1}{4} rad / s^{2}$
Velocity after $10 \sec$ ,
$v = u + at$
$v = 0 + \frac{1}{50} \times 10$
$v = 0.2 m / s$
Angular velocity,
$ω = \frac{v}{r} [∵ v = r ω]$
$ω = \frac{0.2}{8 \times 10^{- 2}} = \frac{20}{8}$
$ω = \frac{5}{2} rad / s$
From law of conservation of energy,
Loss of P.E. $=$ K.E. of translation + K.E. of rotation
$mgh = \frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2}$
$0.4 \times 10 \times 1 = \frac{1}{2} \times 0.4 \times (0.2)^{2} + \frac{1}{2} \times I \times {(\frac{5}{2})}^{2}$
$4 = (0.2 \times 0.04) + \frac{1}{2} \times I \times \frac{25}{4}$
$4 - 0.008 = 3.125 I$
$3.992 = 3.125 I$
$I = \frac{3.992}{3.125}$
$I = 1.277 kg - m^{2}$

AP EAMCET (23.04.2019) Shift-I

Rotational Motion

150163 A thin wire of length $l$ having a linear density $ρ$ is bent into a circular loop with $C$ as its centre as shown in the figure. The moment of inertia of the loop about the line $A B$ is

1

\frac{5}{16} \frac{ρ l^{3}}{π^{3}}

2

\frac{1}{16} \frac{ρ l^{3}}{π^{3}}

3

\frac{1}{8} \frac{ρ l^{3}}{π^{3}}

4

\frac{3}{8} \frac{ρ l^{3}}{π^{2}}

Explanation:

D Given,
Length of thin wire $= l$
Linear density of thin wire $= ρ$
Mass of thin wire, $M) = l ρ$
Circumference $=$ length of thin wire
$2 π R = l$
$R = \frac{l}{2 π}$
original image
The moment of inertia of a thin loop about it's diameter,
$I = \frac{1}{2} {MR}^{2}$
$I = \frac{1}{2} l ρ \times {(\frac{l}{2 π})}^{2}$
$I = \frac{l^{3} ρ}{8 π^{2}}$
Now using parallel axis theorem,
$I_{AB} = I_{C} + {MR}^{2}$
$I_{AB} = \frac{l^{3} ρ}{8 π^{2}} + l ρ \times {(\frac{l}{2 π})}^{2}$
$I_{AB} = \frac{l^{3} ρ}{8 π^{2}} + \frac{ρ l^{3}}{4 π^{2}}$
$I_{AB} = \frac{3}{8} \frac{ρ l^{3}}{π^{2}}$

AP EAMCET (22.04.2019) Shift-I

Rotational Motion

150160 A uniform bar of mass $M$ and length $L$ is bent in the form of an equilateral triangle. Find the moment of inertia of the triangle about an axis passing through the centre of mass and perpendicular to the plane of the triangle

1

{ML}^{2}

2

{ML}^{2} / 2

3

{ML}^{2} / 27

4

{ML}^{2} / 54

Explanation:

D original image
$\tan 30^{\circ} = \frac{x}{L / 6} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{L / 6}$
$x = \frac{L}{6 \sqrt{3}}$
Apply parallel axis theorem,
$I_{Net} = I_{G} + {mx}^{2}$
$= [\frac{1}{12} \times \frac{M}{3} {(\frac{L}{3})}^{2} + \frac{M}{3} {(\frac{L}{6 \sqrt{3}})}^{2}] \times 3$
$= [\frac{M}{36} \times \frac{L^{2}}{9} + \frac{M}{3} \times \frac{L^{2}}{36 \times 3}] \times 3$
$= \frac{M L^{2}}{108} + \frac{M L^{2}}{108}$
$= \frac{2 M L^{2}}{108}$
$I_{N e t} = \frac{M L^{2}}{54}$

J and K-CET-2019

Rotational Motion

150161 A long cylindrical rod is welded to a thin circular disc of diameter $0.5 m$ at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in $m$ is

1

\frac{1}{4}

2

\frac{\sqrt{5}}{8}

3

\frac{1}{2}

4

2 \sqrt{2}

Explanation:

B Given, diameter of disc $= 0.5$ meter
$∴$ Radius of disc $= \frac{0.5}{2} = \frac{1}{4} m$

Moment of inertia of a disc about its diameter,
$I_{CG} = \frac{{MR}^{2}}{4}$
Moment of inertia of a disc about a tangent passing through circumference and parallel to its diameter.
$∴ I = \frac{{MR}^{2}}{4} + {MR}^{2} = \frac{5 {MR}^{2}}{4}$
${MK}^{2} = \frac{5}{4} {MR}^{2} [∵ I = {MK}^{2}]$
$K^{2} = \frac{5}{4} R^{2}$
$K = \sqrt{\frac{5}{4}} \times \frac{0.5}{2}$
$= \frac{\sqrt{5}}{2} \times \frac{1}{4}$
$K = \frac{\sqrt{5}}{8} m$

TS- EAMCET-03.05.2019

Rotational Motion

150162 A wheel of radius $8 cm$ is attached to a support so as to rotate about a horizontal axis through its centre. A string of negligible mass wrapped around its circumference carries a mass of 0.4 $kg$ attached to its free end. When the mass is released, its descends through $1 m$ in 10 seconds, then its moment of inertia is (Acceleration due to gravitv, $g = 10 {m s}^{- 2}$
original image

1

1.277 kg m_{2}^{2}

2

2.177 kg m^{2}

3

21.77 kg m^{2}

4

12.77 kg m^{2}

Explanation:

A Given,
Radius of wheel $(r) = 8 cm = 8 \times 10^{- 2} m$ ,
Mass of body $(m) = 0.4 kg$
Descending length $(s) = 1 m$ , Time $t) = 10 \sec$ .
original image
From second law of motion.
$s = ut + \frac{1}{2} {at}^{2}$
$1 = 0 + \frac{1}{2} a \times (10)^{2}$
$a = \frac{1}{50} m / s^{2}$
Angular acceleration,
$α = \frac{a}{r} [∵ a = α r]$
$α = \frac{\frac{1}{50}}{8 \times 10^{- 2}}$
$α = \frac{100}{50 \times 8}$
$α = \frac{1}{4} rad / s^{2}$
Velocity after $10 \sec$ ,
$v = u + at$
$v = 0 + \frac{1}{50} \times 10$
$v = 0.2 m / s$
Angular velocity,
$ω = \frac{v}{r} [∵ v = r ω]$
$ω = \frac{0.2}{8 \times 10^{- 2}} = \frac{20}{8}$
$ω = \frac{5}{2} rad / s$
From law of conservation of energy,
Loss of P.E. $=$ K.E. of translation + K.E. of rotation
$mgh = \frac{1}{2} {mv}^{2} + \frac{1}{2} I ω^{2}$
$0.4 \times 10 \times 1 = \frac{1}{2} \times 0.4 \times (0.2)^{2} + \frac{1}{2} \times I \times {(\frac{5}{2})}^{2}$
$4 = (0.2 \times 0.04) + \frac{1}{2} \times I \times \frac{25}{4}$
$4 - 0.008 = 3.125 I$
$3.992 = 3.125 I$
$I = \frac{3.992}{3.125}$
$I = 1.277 kg - m^{2}$

AP EAMCET (23.04.2019) Shift-I

Rotational Motion

150163 A thin wire of length $l$ having a linear density $ρ$ is bent into a circular loop with $C$ as its centre as shown in the figure. The moment of inertia of the loop about the line $A B$ is

1

\frac{5}{16} \frac{ρ l^{3}}{π^{3}}

2

\frac{1}{16} \frac{ρ l^{3}}{π^{3}}

3

\frac{1}{8} \frac{ρ l^{3}}{π^{3}}

4

\frac{3}{8} \frac{ρ l^{3}}{π^{2}}

Explanation:

D Given,
Length of thin wire $= l$
Linear density of thin wire $= ρ$
Mass of thin wire, $M) = l ρ$
Circumference $=$ length of thin wire
$2 π R = l$
$R = \frac{l}{2 π}$
original image
The moment of inertia of a thin loop about it's diameter,
$I = \frac{1}{2} {MR}^{2}$
$I = \frac{1}{2} l ρ \times {(\frac{l}{2 π})}^{2}$
$I = \frac{l^{3} ρ}{8 π^{2}}$
Now using parallel axis theorem,
$I_{AB} = I_{C} + {MR}^{2}$
$I_{AB} = \frac{l^{3} ρ}{8 π^{2}} + l ρ \times {(\frac{l}{2 π})}^{2}$
$I_{AB} = \frac{l^{3} ρ}{8 π^{2}} + \frac{ρ l^{3}}{4 π^{2}}$
$I_{AB} = \frac{3}{8} \frac{ρ l^{3}}{π^{2}}$

AP EAMCET (22.04.2019) Shift-I

Question Bank Series

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Question Bank Series

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