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Class 11th Physics

Shown in figure is a block at rest having an inclined smooth groove in vertical plane. The horizontal surface is smooth. A ball of same mass as that of block at rest is released from top end. The time when ball will leave groove is √(nL/(2g√3)). Find n.

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Vishal Baghel

Contributor-Level 10

-mv cos 60? + 2mu = 0 => v = 4u
½m [v² + u² + 2uv cos 120? ] + ½mu² = mgx sin 60?
=> v² = (8/7)√3gx => ar = (4√3/7)g
∴ t = √ (2L * 7)/ (4√3g)

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Class 11th Physics

The given arrangement is in vertical plane and initially the springs are in natural length. The uniform rod is hinged at end O, about which it can freely rotate. The angular frequency of small oscillations of the rod is n/8 √(k/m). Find n. (Given: m = mass of rod, L = length of rod and k = mg/L)

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Vishal Baghel

Contributor-Level 10

Given mg = kL
∴ Iα = (kLθ.L + k (L/2)²θ - mg (L/2)θ)
(mL²/3)α = kL² (3/4)θ (restoring torque)
α = (9k/4m)θ
∴ ω = (3/2)√ (k/m)

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Class 11th Waves

A sine wave is travelling in a medium. The minimum distance between the two particles, always having same speed, is

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Vishal Baghel

Contributor-Level 10

Particles having phase difference of? will move with same speed.

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Class 11th Dual Nature of Radiation and Matter

A free particle having one electronic charge with initial kinetic energy 9 eV and de Broglie wavelength 1 mm enters a region of V₀ potential difference such that new de Broglie wavelength is now 1.5 mm. Then eV₀ is

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Vishal Baghel

Contributor-Level 10

λ? = h/√2mE? = λ? = h/√2mE?
=> E? = (4/9)E? = 4eV
E? = E? - eV? => V? = 5V

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a year ago

Class 11th Mechanical Properties of Solids

A uniform rod, of mass m, length l and r radius of cross section, is rotated about an axis passing through one of its ends and perpendicular to its length with constant angular velocity ω in horizontal plane. If Y is the Young's modulus of the material of rod, the increase in its length due to rotation of rod is

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Vishal Baghel

Contributor-Level 10

dm = (m/L)dx
∴ T = (mω²/2L) (L² - x²)
∴ ΔL = ∫? (mω²/2Lπr²Y) (L² - x²)dx
= ΔL = mω²L²/3πr²Y

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Class 11th Work, Energy and Power

A ball of mass 0.15 kg is dropped from a height. 10 m, strikes the ground and rehounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s² ) nearly:

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alok kumar singh

Contributor-Level 10

Velocity just before striking the ground
v? = √2gh
v? = √ (2*10*10) = 10√2 m/s
v? = -10√2?
If it reaches the same height, speed remains same after collision only the direction changes.
v? = 10√2 m/s
v? = 10√2?
|Impulse| = m|Δv|
= m|10√2? - (-10√2? )|
= 0.15 [2 (10√2)]
= 3√2 kg m/s
= 4.2 kg m/s

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Class 11th Physics

A solid circular disc (m, R) translates with V? and rotates with ω? = 2V₀/R at t = 0. Because of friction, pure rolling starts at t = t₀. The coefficient of friction is μ. Find t₀.

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Vishal Baghel

Contributor-Level 10

a = fr/m
∴ V = V? + at? = V? + (fr/m)t?
and α = 2Rfr/mR² = 2fr/mR
∴ ω = ω? - (2fr/mR)t?
∴ V = ωR ∴ v? + (fr/m)t? = ω? R - (2fr/m)t?
= (3fr/m)t? = V? = (fr/m)et?
∴ V = V? + V? /3 = 4V? /3
t? = mV? / (3μmg)
= V? /3μg

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Class 11th Physics

A particle of mass ' m ' is projected with a velocity v = kVe(k < 1) from the surface of the earth.
(Ve = escape velocity)
The maximum height above the surface reached by the particle is:

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alok kumar singh

Contributor-Level 10

h = R / [ (2gR/V? ²) - 1] = R / [ (V²/K²V? ²) ] = KK² / (1 - K²)

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Class 11th Laws of Motion

A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at
t = 6 s ? (Take g = 10 m/s² )

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alok kumar singh

Contributor-Level 10

velocity of car at t = 4sec is
v = u + at
v = 0 + 5 (4)
= 20 m/s
At t = 6sec
acceleration is due to gravity ∴ a = g = 10 m/s
v? = 20 m/s (due to car)
v? = u + at
= 0 + g (2) (downward)
= 20 m/s (downward)
v = √ (v? ² + v? ²)
= √ (20² + 20²)
= 20√2 m/s

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a year ago

Class 11th Physics System of Particles and Rotational Motion

A uniform rod of length 200 cm and mass 500 g balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 A and another unknown mass 'm' is suspended the rod at 160 cm mark as shown in the Find the value of ' m ' such that the
rod is in equilibrium. (g = 10 m/s²)