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5 months ago

Class 11th Laws of Motion

The boxes of masses 2kg and 8kg are connected by a massless string passing over smooth pulleys. Calculate the time taken by box of mass 8kg to strike the ground starting from rest. (use g = 10 m/s²)

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alok kumar singh

Contributor-Level 10

For 8 kg block,

8g – 2T = 8a . (i)

For 2kg block,

T – 2g = 2 * 2a . (ii)

Solving equations (i) and (ii),

$a = \frac{g}{4} = 2.5 m / s^{2}$

$t = \sqrt[]{\frac{2 s}{a}} = \sqrt[]{\frac{2 * 0.2}{2.5}} = 0.4 s$

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5 months ago

Class 11th Physics Thermodynamics

The ratio of species heats $(\frac{C_{P}}{C_{V}})$ in terms of degree of freedom (f) is given by:

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Raj Pandey

Contributor-Level 9

$\frac{C_{P}}{C_{v}} = y = 1 + 2 / f$

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5 months ago

Class 11th Units and Measurement

Which of the following is not a dimensionless quantity?

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Vishal Baghel

Contributor-Level 10

$μ_{0} = \frac{B_{0}}{H} = N A^{- 2}$

$[μ_{0}] = [M^{1} L^{1} T^{- 2} A^{- 2}]$

It is not dimensionless

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5 months ago

Class 11th System of Particles and Rotational Motion

Moment of inertial of a square plate of side/ about the axis passing through one of the corner and perpendicular to the plane of square plate is given by:

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Vishal Baghel

Contributor-Level 10

$l_{z} = l_{x} + l_{y}$

$l_{z} = \frac{m l^{2}}{3} + \frac{m l^{2}}{3} = \frac{2}{3} m l^{2}$

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5 months ago

Class 11th Motion in a Plane

A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution of its speed is 8 AU per second?

Given : 1 ly = 9.46 * 10¹⁵m

1 AU = 1.5 * 10¹¹m

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Vishal Baghel

Contributor-Level 10

$R = \frac{l}{θ}$

$\Rightarrow T i m e = \frac{4 * 2 π R}{V} = \frac{4 * 2 π x}{V} (\frac{l}{θ})$

Time $= \frac{4 * 2 π * 4.4 * 9.64 * 1 0^{15}}{8 * 1.5 * 1 0^{11} * \frac{4}{3600} * \frac{π}{180}}$

$\Rightarrow T i m e = 4.5 * 1 0^{10} s o c s$

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5 months ago

Class 11th Laws of Motion

The resultant of these forces $\vec{O P}, \vec{O Q}, \vec{Q R}, \vec{O S} a n d \vec{O T}$ is approximately. . . . . . .N

[Take $\sqrt[]{3} = 1.7, \sqrt[]{2} = 1.4 .$ Given $\hat{i} a n d \hat{j}$ unit vectors along x, y axis]

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Vishal Baghel

Contributor-Level 10

${\vec{F}}_{x} = [10 * \frac{\sqrt[]{3}}{2} + 20 * \frac{1}{2} + \frac{20}{\sqrt[]{2}} - \frac{15}{\sqrt[]{2}} - \frac{15 \sqrt[]{3}}{2}] = 9.25 \hat{i}$

${\vec{F}}_{y} = [15 * \frac{1}{2} + 20 * \frac{\sqrt[]{3}}{2} + 10 * \frac{1}{2} - \frac{15}{\sqrt[]{2}} - \frac{20}{\sqrt[]{2}}] = 5 \hat{j}$

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5 months ago

Class 11th Physics

A copper block of mass 5.0kg is heated to a temperature of $500 ° C$ and is placed on a large ice block. What is the maximum amount of ice that can melt?

[Specific heat of copper : 0.39 J g^-1 ⁰C^-1 and latent heat of fusion of water : 335 J g^-1]

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Raj Pandey

Contributor-Level 9

Heat Released by block = Heat gain by large Ice block

$\Rightarrow m c Δ T = M_{i c e} L$

5 * 0.39 * 500 = m_ice * 335

$m_{i c e} = \frac{5 * 0.39 * 500}{335}$

= 2.91 kg

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5 months ago

Class 11th Units and Measurement

Match List II with List II

List I List II

(A) R_H(Rydberg constant) (i) kg m^-¹s^-¹

(B) h (Planck

...more

Match List II with List II

List I List II

(A) R_H(Rydberg constant) (i) kg m^-¹s^-¹

(B) h (Planck's constant) (ii) kg m²s^-¹

(C) $μ_{B}$ (Magnetic field energy density) (iii) m^-¹

(D) $η$ (coefficient of viscocity (iv) kg m^-¹s^-²

less

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alok kumar singh

Contributor-Level 10

R_H : m^-1

h : kgm² s^-1

$μ_{B} : k g m^{- 1} s^{- 2}$

$η : k g m^{- 1} s^{- 1}$

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5 months ago

Class 11th Oscillations

The variation of displacement with time of a particle executing free simple harmonic moton is shown in the figure. The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure:

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Vishal Baghel

Contributor-Level 10

Potential Energy is maximum at extreme position

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5 months ago

Class 11th Physics

A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100kg and radius 50m. If the gravitational potential at a point, 25m from the centre is V kg/m. The value of V is:

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alok kumar singh

Contributor-Level 10

$V_{A} = - \frac{G M_{1}}{r} - \frac{G M_{2}}{R} = - \frac{50 G}{25} - \frac{100 G}{50} = - 4 G$

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