Class 11th

How is the change of state related to latent heat?

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2 months ago

A smooth spherical ball strikes a smooth horizontal floor at an angle $θ = 45^{?}$ . The coefficient of restitution between the ball and the floor is $e = \frac{1}{2}$ . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision)

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Class 11th Physics Thermodynamics

Contributor-Level 9

Vertical component of velocity just after collision $= \frac{u}{2 \sqrt[]{2}}$

$k_{i} = \frac{m u^{2}}{2} k_{f} = \frac{1}{2} m \frac{u^{2}}{2} + \frac{1}{2} m \frac{u^{2}}{8} = \frac{5 m u^{2}}{16}$ $\frac{}{}$

Fraction $= \frac{k_{i} - k_{f}}{k_{i}} = 1 - \frac{\frac{5 m u^{2}}{16}}{\frac{m u^{2}}{2}} = 1 - \frac{5}{8} = \frac{3}{8}$

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Steam at $100^{?} C$ is passed into a calorimeter of water equivalent 10 gm containing 74gm of water and 10gm of ice at . If the temperature of the calorimeter and its contents rises to $0^{?} C$ , the amount of steam passed (Latent heat of vapourisation $= 540 c a l / g$ , latent heat of fusion $= 80 c a l / g$ ) is:

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Class 11th Physics Thermodynamics

Contributor-Level 9

Heat lost $=$ = Heat gained

$\Rightarrow m * 540 + m * s_{ω} * (100 - 5) = 10 * 80 + (10 + 74 + 10) * s_{ω} * 5$

$\Rightarrow m (540 + 95) = 800 + 94 * 5$

$\Rightarrow m 635 = 1270$

$∴ m = 2 g m$

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Three samples of the same gas A,B and $C (γ = 3 / 2)$ have initially equal volume. Now the volume of each sample is doubled. The process is adiabatic for A, isobaric for B and isothermal for C. If the final pressures are equal for all three samples, the ratio of their initial pressures are:

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Contributor-Level 9

For gas $A : P_{A} ($ $_{}$ initial $) = {(\frac{v_{f}}{v_{i}})}^{γ} P = (2)^{3 / 2} P$ ${(\frac{_{}}{_{}})}^{}^{}$

$P_{A} = 2 \sqrt[]{2} P$

For gas B: $P_{B} ($ $_{}$ initial $) = P$

For gas $C : P_{C} ($ initial $) = \frac{2 V_{0}}{V_{0}} P = 2 P ∴$ Ratio $= 2 \sqrt[]{2} : 1 : 2$

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The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is:

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Contributor-Level 10

$η = \frac{W . D . b y g a s}{H e a t a b s o r b}$

$η = \frac{H e a t (N e t)}{H e a t (A b s o r b)}$

$η = \frac{A r e a u n d e r t h e c u r v e}{H e a t A b s o r b}$

$η = \frac{\frac{1}{2} * S_{0} T_{0}}{\frac{1}{2} (3 T_{0}) S_{0}} = \frac{1}{3}$

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Class 11th Waves

A sound source is located somewhere along the x-axis. Experiment shows that the same wavefront simultaneously reaches listeners at x = -10m and x = +4.0m. A third listener is positioned along the positive y-axis. What is y-coordinate (in m) if the same wavefront reaches at him at the same instant it does the first two listeners?

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Class 11th Work, Energy and Power

Contributor-Level 10

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2 months ago

What minimum speed does a 100g particle need At point B to reach point A? The graph shows potential energy versus position.

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Class 11th Work, Energy and Power

Contributor-Level 10

To reach at point A particle must cross the peak point.

Loss in KE = gain in PE

$\frac{1}{2} * \frac{100}{1000} * v^{2} = (5 - 0)$

v² = 100

v = 10 m/s

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Figure shown two blocks A and B having mass 2kg and 4kg moving with a speed 4 m/sec and 2 m/sec respectively. The maximum compression in the spring and final velocity of 2kg block and respectively.

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Contributor-Level 10

$\frac{1}{2} μ v_{r}^{2} = \frac{1}{2} k x^{2}$

$x = {V_{r e r}}^{2} \sqrt[]{\frac{μ}{k}}$

$= 2 * \sqrt[]{\frac{8 / 6}{10}} = \sqrt[]{\frac{8}{15}}$

From conservation of momentum

2 * 4 + 4 * 2 = 2v₁ + 4v₂

8 = v₁ + 2v₂ ….(1)

From conservation of energy

$\frac{1}{2} * 2 * 4^{2} + \frac{1}{2} * 4 * 2^{2} = \frac{1}{2} * 2 * v_{1}^{2} + \frac{1}{2} * 4 * v_{2}^{2}$ ….(ii)

On solving $v_{1} = \frac{4}{3} m / s, v_{2} = \frac{10}{3} m / s$

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2 months ago

A particle is attached to the lower end of a uniform rod which is hinged at its other end as shown in the figure. Another identical particle moving horizontally, collides inelastically and sticks to it. The minimum speed of moving particle so that the rod with particles performs circular motion in a vertical plane will be: [length of rod is $l,$ consider masses of both particles and rod to be same]

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Class 11th Physics Mechanical Properties of Fluids

Contributor-Level 10

From conservation of Angular momentum about hinged point

$m v l = (\frac{m l^{2}}{3} + 2 m . l^{2}) ω$

$ω = (\frac{3}{7} \frac{v}{l})$

Now from conservation of Energy.

$\frac{1}{2} * \frac{7}{3} m l^{2} * ω^{2} = m g l + 2 m g . 2 l$

$v = \sqrt[]{\frac{70}{3} g l}$

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A uniform rod of length $b$ capable of turning about its end which is out of water, rests inclined to the vertical. If the specific gravity of the rod is $\frac{5}{9}$ , find the length of the rod immersed in water.

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