Work, Energy and Power
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In a hydrogen atom, an electron makes a transition from (n + 1)th level to the nth level. If n >> 1, the frequency of radiation emitted is proportional to:
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In a hydrogen atom, an electron makes a transition from (n + 1)th level to the nth level. If n >> 1, the frequency of radiation emitted is proportional to:
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ΔE = have = 13.6Z² (1/n? ² - 1/n? ²)eV.
hv = 13.6 (1²) (1/n² - 1/ (n+1)²) = 13.6 (n+1)²-n²)/ (n² (n+1)²)
v = (13.6/h) * (2n+1)/ (n² (n+1)²). For n>>1, v ≈ (13.6/h) (2n/n? ) ∝ 1/n³.
A small block starts slipping down from a point B on an inclined plane AB, which is making an angle θ with the horizontal section BC. Section BC is smooth and the remaining section CA is rough with a coefficient of friction µ. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by µ = k tanθ. The value of k is
A small block starts slipping down from a point B on an inclined plane AB, which is making an angle θ with the horizontal section BC. Section BC is smooth and the remaining section CA is rough with a coefficient of friction µ. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC = 2AC, the coefficient of friction is given by µ = k tanθ. The value of k is
Let AC = l, BC = 2l, AB = 3l
Apply work-Energy theorem
W? + W? = ΔKE
Mg (3l)sinθ - µmgcosθ (l) = 0+0
µmgcosθl = 3mglsinθ
µ=3tanθ = ktanθ
∴ k=3
A body of mass 2 kg, is driven by an engine delivering, a constant power 1 J/s. the body starts from rest and moves in a straight line. After 9 seconds, the body has moved a distance (in m)
A body of mass 2 kg, is driven by an engine delivering, a constant power 1 J/s. the body starts from rest and moves in a straight line. After 9 seconds, the body has moved a distance (in m)
P = constant
m (dv/dt) = P
∫? vdv = P/m ∫? dt
v²/2 = Pt/m ⇒ v = (2Pt/m)¹/²
dx/dt = √ (2P/m) t¹/²
∫? dx = √ (2P/m) ∫? t¹/² dt
x = √ (2P/m) (t³/² / (3/2) = √ (2P/m) × (2/3) × t³/²
= (2/3) × 27 = 18
A particle of mass is fixed to one end of a light spring having force constant and unstretched length . The other end is fixed. The system is given an angular speed about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:
A particle of mass is fixed to one end of a light spring having force constant and unstretched length . The other end is fixed. The system is given an angular speed about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:
A ball of mass 10 kg moving with a velocity10√3ms-1along X – axis, hits another ball of mass 20 kg which is at rest. After collision, the first ball comes to rest and the second one disintegrates into two equal pieces. One of the pieces starts moving along Y– axis at a speed of 10m/s. The second piece starts moving at a speed of 20m/s at an angle θ° with respect to the X –axis. The configuration of pieces after collision is shown in the figure. The value of θ to the nearest integer is----------.
A ball of mass 10 kg moving with a velocity10√3ms-1along X – axis, hits another ball of mass 20 kg which is at rest. After collision, the first ball comes to rest and the second one disintegrates into two equal pieces. One of the pieces starts moving along Y– axis at a speed of 10m/s. The second piece starts moving at a speed of 20m/s at an angle θ° with respect to the X –axis. The configuration of pieces after collision is shown in the figure. The value of θ to the nearest integer is----------.
A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance h = 9.8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is: (take g = 9.8ms?²)
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A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance h = 9.8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is: (take g = 9.8ms?²)
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Using conservation of linear momentum, we can write
P? = P? ⇒ mv = (M+m)V
Using conservation of Mechanical energy, we can write
½ (M+m)V² = (M+m)gh ⇒ V = √2gh
⇒ v = (M+m)/m)√2gh
= (6/0.01)√ (2×9.8×0.098) = 600 × 1.386 = 831.4 m/s
A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation F = F₀[1 - ((t-T)/T)²]
Where F₀ and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is:
A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation F = F₀[1 - ((t-T)/T)²]
Where F₀ and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is:
v = ∫a dt = (1/M)∫F dt
v = (F? /M) ∫ [1 - (t-T)²/T²] dt from 0 to 2T
= (F? /M) [t - (t-T)³/3T²]? ²?
= (F? /M) [ (2T - T³/3T²) - (0 - (-T)³/3T²) ]
= (F? /M) [ 2T - T/3 - T/3 ] = (F? /M) [ 4T/3 ] = 4F? T/3M.



