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Physics System of Particles and Rotational Motion

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Physics System of Particles and Rotational Motion Class 11th

A uniform thin bar of mass 6kg and length 24 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane hexagon is………….* 10^-1 kg m².

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Vishal Baghel

Contributor-Level 10

Let m_AB = m = 1 kg

AB = 0.4 m = $l$

d = OM = $\sqrt[]{0.16 - 0.04} = \sqrt[]{0.12}$

$d = 2 \sqrt[]{3} * 1 0^{- 1}$

$I_{A B a b o u t O} = \frac{m l^{2}}{12} + m d^{2}$

$∴ l_{h e x a g o n, a b o u t O} = 6 [\frac{m l^{2}}{12} + m d^{2}] = \frac{m l^{2}}{2} + 6 m d^{2}$

$= \frac{1 * {(0.4)}^{2}}{2} + 6 * 1 * {(2 \sqrt[]{3} * 1 0^{- 1})}^{2} = \frac{0.16}{2} + 6 * 4 * 3 * 1 0^{- 2}$

= 0.08 + 0.72 = 0.8 kgm² = 8 * 10^-¹ kgm²

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Physics System of Particles and Rotational Motion Class 11th

A circular hole of radius $(\frac{a}{2})$ is cut of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be:

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Vishal Baghel

Contributor-Level 10

$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$

$= \frac{σ π a^{2} * a - \frac{σ π a^{2}}{4} * \frac{3 a}{2}}{σ π a^{2} - \frac{σ π a^{2}}{4}} = \frac{\frac{5}{8} σ π a^{3}}{\frac{3}{4} σ π a^{2}} = \frac{5}{8} σ π a^{2} * \frac{4}{3} σ π a^{2} = \frac{5}{6} a$

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Physics System of Particles and Rotational Motion Class 11th

Match List-I with List-II.

List-I List-II

(Material) &nb

...more

Match List-I with List-II.

List-I List-II

(Material) (Susceptibility ( $χ$ ))

A. Diamagnetic I. $χ = 0$

B. Ferromagnetic II. $0 > x \geq - 1$

C. Paramagnetic III. $χ ? 1$

D. Non-magnetic IV. $0 < χ < ε$ (a small positive number)

Choose the correct answer from the options given below

less

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alok kumar singh

Contributor-Level 10

(Material) (Susceptibility ( $χ$ )

Diamagnetic (II) $0 > χ \geq - 1$

Ferromagnetic (III) $χ ? 1$

Paramagnetic (IV) $0 < χ < ε$

Non-magnetic (I) $χ = 0$

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2 months ago

Physics System of Particles and Rotational Motion Class 11th

A light ray enters through a right angled prism at point $P$ with the angle of incidence $3 0^{\circ}$ as shown in figure. It travels through the prism parallel to its base $B C$ and emerges along the face $A C$ . The refractive index of the prism is:

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alok kumar singh

Contributor-Level 10

$A = 9 0^{?}$

In prism, $r_{1} + c = A$

$r_{1} = 9 0^{?} - c$ …(i)

$sin c = \frac{1}{μ} \Rightarrow cos c = \frac{\sqrt[]{μ^{2} - 1}}{μ}$

$\Rightarrow$ Apply Snell's law, on incidence surface

$1 \cdot sin 3 0^{?} = μ sin (r_{1}) \Rightarrow 1 * \frac{1}{2} = μ * sin (9 0^{?} - c)$

$\frac{1}{2} = μ * \frac{\sqrt[]{μ^{2} - 1}}{μ}$

On squaring $\frac{1}{4} = μ^{2} - 1$

$\Rightarrow μ^{2} = \frac{5}{4} \Rightarrow μ = \frac{\sqrt[]{5}}{2}$

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2 months ago

Physics System of Particles and Rotational Motion Class 11th

A 2kg steel rod of length 0.6m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the firction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is________ms^-1.

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alok kumar singh

Contributor-Level 10

$l = \frac{m l^{2}}{3}$

Energy conservation Low

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

And speed V = $ω r = ω l$

then $V = \sqrt[]{6 g l} = \sqrt[]{6 * 10 * . 6}$

->6 m/s

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2 months ago

Physics System of Particles and Rotational Motion Class 11th

A disc of mass 1kg and radius R is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $\sqrt[]{\frac{x}{3 R} r a d s^{- 1}}$ where x = __________.

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Payal Gupta

Contributor-Level 10

Loss in P.E. = Gain in k.E

2 mg R = $\frac{1}{2} (\frac{1}{2} m R^{2} + m R^{2}) ω^{2}$

$ω = \sqrt[]{\frac{8 g}{3 R}} = 4 \sqrt[]{\frac{g}{2 * 3 R}}$

$x = \frac{g}{2} = 5$

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2 months ago

Physics System of Particles and Rotational Motion Class 11th

The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is $2400 g c m^{2}$ . The length of the 400 g rod is nearly:

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alok kumar singh

Contributor-Level 10

Moment of inertia of $r o d = I = \frac{m l^{2}}{12}$

$\Rightarrow 2400 = 400 \frac{l^{2}}{12}$

$\Rightarrow 72 = l^{2}$

$\Rightarrow l = \sqrt[]{72} = 8.48 cm = 8.5 cm$

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2 months ago

Physics System of Particles and Rotational Motion Class 11th

A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ( $P$ and $Q$ are any highest and lowest points on the wheel, respectively)?

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alok kumar singh

Contributor-Level 10

In the case of pure rolling,

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Physics System of Particles and Rotational Motion System of Particles and Rotational Motion

The centre of a wheel rolling on a plane surface moves with a speed $υ_{0} .$ A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt[]{x} υ_{0} .$ Then the value of x is__________.

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Vishal Baghel

Contributor-Level 10

$V_{P} = \sqrt[]{V_{0}^{2} + V_{0}^{2}} = \sqrt[]{2} V_{0}$

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Physics System of Particles and Rotational Motion Class 11th

The position of the centre of mass of a uniform semi – circular wire of radius ‘R’ placed in x – y plane with its centre at the origin and the line joining its ends as x-axis is given by $(0, \frac{x R}{π}) .$ Then, the value of $| x |$ is___________.

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