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Class 11th Physics

The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $ρ (r) = ρ_{0} (1 - \frac{r^{2}}{R^{2}})$ . Then the gravitational field is maximum at:

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alok kumar singh

Contributor-Level 10

$E = \frac{G m_{enc}}{r^{2}} = \frac{G}{r^{2}} \int_{0}^{r} ? ρ_{0} (1 - \frac{r^{2}}{R^{2}}) 4 π r^{2} d r$

$\begin{array}{r} E & = \frac{G}{r^{2}} 4 π ρ_{0} {[\frac{r^{3}}{3} - \frac{r^{5}}{5 R^{2}}]}_{0}^{r} \\ = 4 π G ρ_{0} [\frac{r}{3} - \frac{r^{3}}{5 R^{2}}] \end{array}$

For $E$ to be $m a x, \frac{d E}{d r} = 0$

$\begin{matrix} \Rightarrow \frac{1}{3} - \frac{3 r^{2}}{5 R^{2}} = 0 \\ r = \sqrt[]{\frac{5}{9}} R \end{matrix}$

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Class 11th Thermodynamics

To raise the temperature of a certain mass of gas by $50^{\circ} C$ at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by $100^{\circ} C$ at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

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alok kumar singh

Contributor-Level 10

$Δ Q = {n C}_{p} Δ T \Rightarrow 160 = {n C}_{p} 50$

$Δ Q = {n C}_{v} Δ T \Rightarrow 240 = {n C}_{v} 100$

$\begin{array}{r} \Rightarrow \frac{C_{p}}{C_{v}} = \frac{16}{5} * \frac{10}{24} = \frac{4}{3} \\ \Rightarrow 1 + \frac{2}{f} = \frac{4}{3} \Rightarrow f = 6 \end{array}$

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Class 11th Thermodynamics

The specific heat of water $= 4200 J {k g}^{- 1} K^{- 1}$ $^{}^{}$ and the latent heat of ice $= 3.4 * 10^{5} J {k g}^{- 1}$ $^{}^{}$ . 100 grams of ice at $0^{?} C$ $^{}$ is placed in $200 g$ of water at $25^{?} C$ $^{}$ . The amount of ice that will melt as the temperature of water reaches $0^{?} C$ $^{}$ is close to (in grams):

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Vishal Baghel

Contributor-Level 10

$m (L) = m_{1} S_{1} (Δ T)$

$\Rightarrow m (3.4 * 10^{5}) = (200) (4200) (25)$

$\Rightarrow m = 61.7$

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Class 11th Thermodynamics

A metallic sphere cools from $50^{\circ} C$ to $40^{\circ}$ in 300 s. If atmospheric temperature around is $20^{\circ} C$ , then the sphere's temperature after the next 5 minutes will be close to:

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alok kumar singh

Contributor-Level 10

$\frac{50 - 40}{300} = - k [\frac{50 + 40}{2} - 20]$

$\begin{array}{r} \frac{40 - T}{300} - k [\frac{T + 40}{2} - 20] \\ \Rightarrow \frac{10}{40 - T} = \frac{25}{T} * 2 \\ \Rightarrow T = 200 - 5 T \\ \Rightarrow T = \frac{100}{3} \approx 33^{?} C \end{array}$

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Class 11th Oscillation

A block of mass $m$ attached to a massless spring is performing oscillatory motion of amplitude ' $A$ ' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become $f A$ . The value of $f$ is:

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alok kumar singh

Contributor-Level 10

$V_{m a x} = A^{'} ω^{'} = A ω$

$\begin{array}{r} A^{'} \sqrt[]{\frac{k}{m / 2}} = A \sqrt[]{\frac{k}{m}} \\ A^{'} = \frac{A}{\sqrt[]{2}} \end{array}$

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Class 11th Physics

Dimensional formula for thermal conductivity is (here K denotes the temperature):

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Vishal Baghel

Contributor-Level 10

$K = \frac{(\frac{Q}{r}) Δ x}{Δ A T}$

$\begin{array}{l} \Rightarrow & \frac{({M L}^{2} T^{- 2}) (L)}{(L^{2}) (θ) (T)} \\ \Rightarrow & M^{1} L^{1 -} T^{- 3} θ^{- 1} \end{array}$

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Class 11th Physics

A Tennis ball is released from a height $h$ and after freely falling on a wooden floor it rebounds and reaches height $\frac{h}{2}$ $\frac{}{}$ . The velocity versus height of the ball during its motion may be represented graphically by:

(graphs are drawn schematically and on not to scale)

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Vishal Baghel

Contributor-Level 10

$\begin{array}{r} \Rightarrow V^{2} = U^{2} + 2 g S \\ \Rightarrow V^{2} = 0 + 2 g (h - y) \\ \Rightarrow V^{2} = 2 g h - 2 g y \\ \Rightarrow V = \sqrt[]{2 g h - 2 g y} \end{array}$

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Class 11th Physics

A calorimeter of water equivalent $20 g$ contains $180 g$ of water at $25^{?} C$ , ' $m$ ' grams of steam at $100^{?} C$ is mixed in it till the temperature of the mixture is $31^{?} C$ . The value of

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alok kumar singh

Contributor-Level 10

Heat lost by steam $=$ Heat gained by water and calorimeter.

$\begin{array}{r} m * 540 + m * 1 * (100 - 31) = 200 \\ 540 m + 69 m = 1200 \\ m = \frac{1200}{609} \approx 2 \end{array}$

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Class 11th Units and Measurement

Amount of solar energy received on the earth's surface per unit area per unit time is defined a solar constant. Dimension of solar constant is:

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alok kumar singh

Contributor-Level 10

Solar constant $= \frac{Energy}{Area * Time} = \frac{{M L}^{2} T^{- 2}}{L^{2} * T} = {M L}^{?} T^{- 3}$

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Class 11th Physics System of Particles and Rotational Motion

A uniform rod of length 'l' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $ω$ the rod makes an angle $θ$ with it (see figure). To find $θ$ equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} ω^{2} s i n ? θ c o s ? θ$ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $C M$ . The value of $θ$ is then such that: