Physics

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%.

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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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5 months ago

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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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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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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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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5 months ago

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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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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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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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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5 months ago

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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Physics Physics Mechanical Properties of Fluids

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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5 months ago

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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Physics Work, Energy and Power

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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5 months ago

A bag of sand of mass 9.8kg is suspended by a rope. A bullet of 200g travelling with speed 10 ms^-1 gets embedded in it, then loss of kinetic energy will be

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Contributor-Level 10

$v = \frac{m v_{0}}{M + m} \Rightarrow K_{f} = \frac{1}{2} (M + m) v^{2} = \frac{m v_{0}^{2}}{2 (M + m)}$

$K_{i} = \frac{m v_{0}^{2}}{2}$

Loss in Kinetic energy = $Δ K = K_{i} - K_{f}$

$Δ K = \frac{1}{2} m v_{0}^{2} (\frac{M}{M + m}) = \frac{1}{2} * 0.2 * 100 = [\frac{9.8}{10}] = 9.8 J$

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5 months ago

Physics Physics Gravitation

An object is taken to a height above the surface of earth at a distance $\frac{5}{4} R$ from the centre of the earth. Where radius of earth, R = 6400km. The percentage decrease in the weight of the object will be

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Contributor-Level 10

As we know that $g_{P} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{\frac{5}{4} R})}^{2} = \frac{16 g}{25}$

$\Rightarrow Δ g = g - g_{P} = \frac{9 g}{25}$

The percentage decrease in the weight of the object = $\frac{Δ g}{g} * 100 = 36 %$

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5 months ago

Let $η_{1}$ is the efficiency of an engine at T₁ = $447 ° C$ and T₂ = $147 ° C$ while $η_{2}$ is the efficiency at T₁ = $947 ° C$ and T₂ $47 ° C$ . The ratio $\frac{η_{1}}{η_{2}}$ will be

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