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11 months ago

Physics Class 11th

What will be the effect on the root mean square velocity of oxygen molecules if the temperagture is doubled and oxygen molecule dissociates into atomic oxygen?

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alok kumar singh

Contributor-Level 10

V_rms = $\sqrt[]{\frac{3 R T}{M}}$

V_rms' = $\sqrt[]{\frac{3 R T_{f}}{m_{f}}} = \sqrt[]{\frac{3 R * 2 T * 2}{M}}$ $[\begin{matrix} G i v e n, T_{f} = 2 T & M_{f} = \frac{M}{2} \end{matrix}]$

$= 2 \sqrt[]{\frac{3 R T}{M}}$

V_rms' = 2V_rms

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11 months ago

Physics Class 11th

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:

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alok kumar singh

Contributor-Level 10

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B

...more

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B to C

Q_BC = $Δ u + w_{B C}$

Þ 0 = U_c – U_D – 50

U_C = U_B + 50

= 1600 + 50

= 1650 J

U_A – U_C = $Δ U$ (Path C to A) = 1560-1650

= -90 J

1st Low for path C to A

Q_CA = $Δ U_{C A} + w_{C A} + w_{C A}$

Þ -60 = -90 + w_CA

W_CA = +30J

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11 months ago

Physics Physics Moving Charges and Magnetism

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11 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Ten

A water drop of radius $1 μ m$ falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 * 10^-⁵ Nsm^-² and its density is negligible as compared to that of water 10⁶ gm^-³. Terminal velocity of the water drop is:

(Take acceleration due to gravity = 10 ms^-²)

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alok kumar singh

Contributor-Level 10

NLM 1

$ρ_{ω} \frac{4}{3} π r^{3} g = 6 π η r V_{T}$

$\frac{ρ_{ω} 4}{3} π r^{2} g =$ $6 π η v_{T}$

$V_{T} = \frac{4}{3} \frac{ρ_{ω} π r^{2} g}{6 π η}$

$= \frac{2}{9} * 1 0^{3} * \frac{{(1 0^{- 6})}^{2} * 10}{1.8 * 1 0^{- 5}}$

$= 0.1234 * 1 0^{- 3}$

vT = 123.4 * 10^-6m/sec

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11 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eight

Two objects of equal masses placed at certain from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other, then the new force will be:

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alok kumar singh

Contributor-Level 10

F = $\frac{G m_{2}}{d^{2}}$ - (i)

Now, $F' = \frac{G m_{1} m_{2}}{d^{2}} = \frac{G 2 m}{3} * \frac{4 m}{3 d^{2}}$

$k' = \frac{8}{9} G \frac{m^{2}}{d^{2}}$

$∴ F' = \frac{8}{G} F$

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11 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Six

Water falls from a 40m height dam at the rate of 9 * 10⁴ kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydro electric energy number of 100 W lamps, that can be lit, is:

(Take g = 10 ms^-2)

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alok kumar singh

Contributor-Level 10

Given : water flow rate = $(\overset{o}{m})$ = 9 * 10⁴kg/hr

= $\frac{9 * 1 0^{4} k g / s e c}{60 * 60}$

Potential energy available on water per unit time

$\Rightarrow \overset{o}{m} g h = \frac{9 * 1 0^{4}}{60 * 60} * 10 * 40$

$\overset{o}{m} g h = 1 0^{4} w a t t$

50% of it converts into electrical energy = 50% of $\overset{o}{m} g h = \frac{1 0^{4}}{2} w a t t$

Let $η$ be the number of bulbs n = 100 = $\frac{1 0^{4}}{2}$

n = 50

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11 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A $\sqrt[]{34} m$ long ladder weighing 10kg leans on a frictionless wall. Its feet rest on the floor 3m away from the wall as shown in the figure. If F_f and F_w are the reaction forces of the floor and the wall, then ratio of F_w / F_f will be:

(Use g = 10 m/s²)

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alok kumar singh

Contributor-Level 10

mg = 10 * 10

= 100 of

NA = $F_{ω}$ reaction force offered by the wall

$\sqrt[]{N_{3}^{2} + f_{B}^{2}} = F_{f}$ reaction force offered by the floor.

By NLM 1

Translational fB = NA

NB = mg = 100N

Ac² + Bc² = AB²

Ac² = 34 – 9

Ac = 5m

Rotational equilibrium $τ_{B} = 0$

mg $\frac{l}{2} c o s θ = N_{A} l s i n θ$

$100 \frac{c o s θ}{2} = N_{A} s i n θ$

NA = 50 cot

Now cot = $\frac{3}{5}$

N_A= $50 * \frac{3}{5} = 30 N$

Therefore, NA = $F_{ω} = 30 N$ - (i)

$F_{f} = \sqrt[]{N_{B}^{2} + f_{B}^{2}} = \sqrt[]{10 0^{2} + 3 0^{2}} = 10 \sqrt[]{109} N$ - (2)

$\frac{f_{ω}}{F_{f}} = \frac{30}{10 \sqrt[]{109}} = \frac{3}{109}$

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11 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Six

A block of mass 2kg moving on a horizontal surface with speed of 4 ms^-¹ enters a rough surface ranging from x = 0.5m to x = 1.5m. The retarding force in this range of rough surface is related to distance by F = -kx where k = 12Nm^-¹. The speed of the block as it just crosses the rough surface will be:

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alok kumar singh

Contributor-Level 10

by, WET $\int F d x = Δ k . 6$

$\int - k x d x = \frac{1}{2} m [v_{t}^{2} - v_{i}^{2}]$

$\Rightarrow \frac{- k}{2} {[x_{f}^{2} - x_{i}^{2}]}_{0.5}^{1.5} = \frac{1}{2} * 2 [v_{f}^{2} - 16]$ $[? v_{i} = 4 m / s e c]$

$\frac{- 12}{2} [1 . 5^{2} - 0 . 5^{2}] = [v_{f}^{2} - 16]$

$v_{f}^{2} = 16 - 12 = 4$

vf = 2m/sec

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11 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A ball is spun with angular acceleration a = 6t² – 2t where t is in second and a is in rads^-¹. At t = 0, the ball has angular velocity of 10 rads^-¹ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

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alok kumar singh

Contributor-Level 10

a = 6t² – 2t, of t 20, w = 10 rad/sec

, q = 4 rad

$α = \frac{d ω}{d t}$

$\Rightarrow \int_{10}^{ω} d ω = \int_{t - 0}^{t} α d t$

$\Rightarrow ω - 10 = \int_{0}^{t} (6 t^{2} - 2 t) d t$

$ω = \frac{d θ}{d t}$

$d θ = ω d t$

$\int_{4}^{θ} d θ = \int_{0}^{t} (2 t^{3} - t^{2} + 10) d t$

= $\frac{t^{4}}{2} - \frac{t^{3}}{3} + 10 t + 4$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

Velocity $(υ)$ and acceleration (a) in two system of units 1 and 2 are related as $υ_{2} = \frac{n}{m^{2}} υ_{1} a n d a_{2} = \frac{a_{1}}{m n}$ respectively. Here m and n are constants. The relations for distance and time in two system respectively are:

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alok kumar singh

Contributor-Level 10

Let, L₁ & L₂ one distance for system (1) & (2) respectively

T₁ & T₂ are time for system (1) & (2) respectively

Given : v2 = $\frac{n}{m^{2}} v_{1}$

$\Rightarrow (\frac{L_{2}}{T_{2}}) = \frac{n}{m_{2}} (\frac{4}{T_{1}})$ $[? v_{2} = \frac{L_{2}}{T_{2}}, v_{1} = \frac{L_{1}}{T_{1}}]$

$\Rightarrow \frac{L_{2}}{L_{1}} = (\frac{n}{m_{2}}) [\frac{T_{2}}{T_{1}}] - - (i)$

And a2 = $\frac{a_{1}}{m n} \Rightarrow \frac{v_{2}}{T_{2}} = \frac{v_{1}}{T_{1} m n}$

$\Rightarrow \frac{T_{1}}{T_{2}} = \frac{v_{1}}{v_{2}} (\frac{1}{m n})$ $[? \frac{v_{1}}{v_{2}} = \frac{m^{2}}{n}]$

$\Rightarrow \frac{n^{2}}{m} T_{1} = T_{2}$

from (1)

$\frac{L^{2}}{L_{1}} = \frac{n}{m^{2}} [\frac{n^{2}}{m}] [? \frac{T_{2}}{T_{1}} = \frac{n^{2}}{m}]$

$\Rightarrow \frac{n^{3}}{m^{3}} L_{1} = L_{2}$

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