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11 months ago

Physics Physics Current Electricity

Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are r₁ and r₂ (r₁ > r₂). If the potential difference across the source of internal resistance r₁ is zero, then the value of R will be:

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alok kumar singh

Contributor-Level 10

V_AB = 0

$o r, ε - i r_{1} = 0$

$\Rightarrow ε - \frac{2 ε}{r_{1} + r_{2} + R} r_{1} = 0 \Rightarrow 1 = \frac{2 r_{1}}{r_{1} + r_{2} + R}$

$\Rightarrow R + r_{1} + r_{2} = 2 r_{1}$

$R = r_{1} - r_{2}$

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11 months ago

Physics Class 12th

Two identical positive charge Q each are fixed at a distance of '2a' apart from each other. Another point charge q₀ with mass 'm' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge q₀ executes SHM. The time period of oscillation of charge q₀ will be:

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alok kumar singh

Contributor-Level 10

$F_{R} = {\frac{k Q q_{0}}{{(a - x)}^{2}} - \frac{k Q q_{0}}{{(a + x)}^{2}}}$

$= - k Q q_{0} {\frac{a^{2} + x^{2} + 2 a x - a^{2} - x^{2} + 2 a x}{(a^{2} - x^{2})}}$

$F_{R} = - \frac{k Q q_{0} (4 a x)}{{(a^{2} - x^{2})}^{2}}$

$T = 2 π \sqrt[]{\frac{a^{3} m}{4 k Q q_{0}} = \sqrt[]{\frac{4 π^{2} a^{3} m * 4 π ε_{0}}{4 Q q_{0}}}} = \sqrt[]{\frac{4 π^{3} ε_{0} m a^{3}}{Q q_{0}}}$

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Physics Class 11th

Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is 1 : 4, then

(A) The r.m.s. velocity of gas molecules in two vessels will be the same.

(B) The ratio of pressure in these vessels will be 1 : 4.

(C) The ratio of pressure will be 1 : 1.

(D) The r.m.s. velocity of gas molecules in two vessels will be 1 : 4.

Choose the correct answer form the options given below:

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alok kumar singh

Contributor-Level 10

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} A s . v_{1} = v_{2}$

$V_{r m s} α \sqrt[]{T}$ $η_{2} = 4 n_{1}$

$T_{1} = T_{2} \Rightarrow V_{r m s - 1} = V_{r m s - 2}$

$P_{1} = \frac{n_{1} R T}{v} P_{z} = \frac{n_{2} R T}{v}$

$\frac{P_{1}}{P_{2}} = \frac{n_{1}}{n_{2}} = \frac{1}{4}$

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11 months ago

Physics Class 11th

Read the following statements:

(A) When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice.

(B) Two bodies P and Q having equal surface areas are maintained at temperature 10°C and 20°C. The thermal radiation emitted in a given time by P and Q are in the ratio 1:1. 15.

(C) A Carnot Engine working between 100K and 400K has an efficiency of 75%.

...more

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alok kumar singh

Contributor-Level 10

A.A is correct statement

$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{100}{400} = \frac{3}{4} = 75 %$

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11 months ago

Physics Class 11th

If K₁ and K₂ are the thermal conductivities, L₁ and L₂ are the lengths and A₁ and A₂ are the cross sectional areas of steel and copper rods respectively such that $\frac{K_{2}}{K_{1}} = 9, \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2 .$ Then, for the arrangement as shown in the figure, the value of temperature T of the steel – copper junction in the steady state will be:

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alok kumar singh

Contributor-Level 10

$\frac{K_{2}}{k_{1}} = 9 \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2$

$\frac{450 - T}{\frac{L_{1}}{A_{1} k_{1}}} = \frac{T}{\frac{L_{2}}{k_{2} A_{2}}} \Rightarrow (450 - T) \frac{L_{2}}{K_{2} A_{2}} = (T) \frac{L_{1}}{A_{1} k_{1}}$

$(450 - T) = T (\frac{L_{1}}{L_{2}} \frac{k_{2} A_{2}}{k_{1} A_{1}})$ $\Rightarrow 450 - T = T (2 * 9 * \frac{1}{2}) = 9 T$

10 T = 450

T = 45° C

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11 months ago

Physics Physics Atoms

Hydrogen atom from exited state comes to the ground state by emitting a photon of wavelength $λ .$ The value of principal quantum number 'n' of the excited state will be

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Payal Gupta

Contributor-Level 10

According to definition of wave number, we can write

$\frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{n^{2}}) \Rightarrow 1 - \frac{1}{n^{2}} = \frac{1}{λ R}$

$\Rightarrow \frac{1}{n^{2}} = 1 - \frac{1}{λ R} = \frac{λ R - 1}{λ R} \Rightarrow n = \sqrt[]{\frac{λ R}{λ R - 1}}$

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11 months ago

Physics Class 11th

Two satellites A and B, having masses in the ration 4 : 3, are revolving in circular orbits of radii 3r and 4r respectively around the earth. The ratio of total mechanical energy of A to B is:

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alok kumar singh

Contributor-Level 10

$\frac{E_{A}}{E_{B}} = \frac{- \frac{G M m_{A}}{2 * 3 r}}{\frac{- G M m_{B}}{2 * 4 r}} = \frac{m_{A}}{m_{B}} * \frac{4}{3} = \frac{4}{3} * \frac{4}{3} = \frac{16}{9}$

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11 months ago

Physics Class 12th

The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1%, 2% and 3% respectively. The maximum percentage error in the detection of the dissipated heat will be

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Payal Gupta

Contributor-Level 10

$H = l^{2} R t$

$% e r r o r i n H = \frac{Δ H}{H} * 100 = 2 (\frac{Δ l}{l}) * 100 + (\frac{Δ R}{R}) * 100 + (\frac{Δ t}{t}) * 100 = 2 * 2 + 1 + 3 = 8 %$

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11 months ago

Physics Class 11th

A ball is projected from the ground with a speed 15 ms^-1 at an angle θ with horizontal so that its range and maximum height are equal. Then 'tanθ' will be equal to

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Payal Gupta

Contributor-Level 10

$H = \frac{u^{2} s i n^{2} θ}{2 g} a n d R = \frac{u^{2} s i n 2 θ}{g}$

According to question

$R = H \Rightarrow \frac{u^{2} s i n^{2} θ}{2 g} = \frac{2 u^{2} s n θ c o s θ}{g} \Rightarrow t a n θ = 4$

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11 months ago

Physics Physics Mechanical Properties of Fluids

Two cylindrical of equal cross-sectional are 16 cm² contain water upto height 100cm and 150 cm respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, it [Take, density of water = 10³ kg/m³ and g = 10 ms^-2]:

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alok kumar singh

Contributor-Level 10

$ρ = 1 0^{3} k g / m^{3}$

g = 10 m/s²

Final height in both vessels

$= \frac{100 + 150}{2} = 125 c m$

So, less in U = $\int A * 0.25 * g * 0.25$

$1 0^{3} * 625 * 1 0^{- 4} * 10 * 16 * 1 0^{- 4}$

= 625 * 16 * 104

= 1J.

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