Physics Test - 16

Given:

\(I = \frac{Q}{t} =\frac{96000}{3600}=\frac{80}{3A}\)

\(V =50\) V

\(t =1\) hour \(=3600\) sec

Where, \(I =\) current, \(R =\) resistance, \(t =\) the time taken, \(V =\) electric potential and \(Q =\) quantity of charge flowing

We know that:

The amount of heat produced (H) in joules is:

\(H = Vlt\)

\(\Rightarrow H=50 \times \frac{80}{3} \times 3600\)

\(\Rightarrow H =4.8 \times 10^{6} J\)

From Newton's law of viscosity

\(\tau=\mu \frac{d u}{d y}\)

We know that

\(F =\tau \times A\)

\(\therefore F=\mu A \frac{d u}{d y}\)

\(\therefore \frac{F_{2}}{F_{1}}=\frac{u_{2}}{u_{1}}\)

Given: \(F_{2}=3 kN , F_{1}=0.9 kN , u _{1}=1.25 cm / s\)

\(\therefore \frac{F_{2}}{F_{1}}=\frac{u_{2}}{u_{1}}\)

\(\therefore \frac{3}{0.9}=\frac{u_{2}}{1.25}\)

\(\therefore u _{2}=4.166 cm / s\)

Given,

\(\lambda_{1}=3 \lambda_{2}\)

According to Einsteins photoelectric equation

\(\frac{h c}{\lambda}=\phi_{0}+K E_{\max }\)

\(\therefore K_{1}=\frac{h c}{\lambda_{1}}-\phi_{0}\) and \(K_{2}=\frac{h c}{\lambda_{2}}-\phi_{0}\)

or,

\(K_{1}-K_{2}=h c\left[\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right]\)

\(=h c\left[\frac{1}{3 \lambda_{2}}-\frac{1}{\lambda_{2}}\right]=-\frac{2 h c}{3 \lambda_{2}}\)

\(\Rightarrow K_{1}-K_{2}=-\frac{2}{3}\left(K_{2}+\phi_{0}\right)\)

or,

\(K_{1}=K_{2}-\frac{2}{3} \phi_{0}=\frac{K_{2}}{3}-\frac{2}{3} \phi_{0}\)

Or,

\(K_{1}<\frac{K_{2}}{3}\)

\(\Rightarrow K_{1}=3 K_{2}\)

Given:

\(P _{1}=200 W\)

\(P _{2}=100 W\)

\(V =100 V\)

We know that:

The rate at which the electric energy is dissipated or consumed is termed as electric power.

The electric power is given as,

\(P=I V=I^{2} R=\frac{V^{2}}{R}\)

Where, \(P =\) electric power, \(V =\) voltage, \(I =\) current and \(R =\) resistance

\(R=\frac{V^{2}}{P}\)

\(R \propto \frac{1}{P} \quad \ldots(1)\)

For an electric bulb, the resistance of the bulb is inversely proportional to the power of the bulb.

So, the bulb which has more power will have low resistance, therefore the \(100 W\) bulb will have more resistance compared to the \(200 W\) bulb.

The heat dissipated by the bulb is given as,

\(H = I ^{2} R\)

Both the bulbs are connected in series so the current in both the bulbs will be equal. So the heat dissipated will be more in the bulb which has more resistance.

Since the resistance of the \(100 W\) bulb is more, so the heat dissipation of the \(100 W\) bulb will be more.

The brightness of the bulb depends on the heat dissipation by the bulb, so the \(100 W\) bulb will have more brightness.

The excess pressure inside the soap bubble is given by,

\(\Delta P=\frac{4 \sigma}{R}\)

where \(\sigma=\) Surface tension, \(R=\) Radius of bubble

\(D =10 mm , R =5 mm =0.005 m \sigma=0.04 N / m\)

\(\Delta P=\frac{4 \sigma}{R}=\frac{4 \times 0.04}{0.005}=32 Pa\)

Given masses of three liquids are equal,

i.e. \(m_x=m_y=m_z=m\)

Temperature of mixture of liquids \(x\) and \(y, \mathrm{~T}_1=16^{\circ} \mathrm{C}\)

Temperature of mixture of liquids \(y\) and \(z, T_2=26^{\circ} \mathrm{C}\)

While mixing the liquids \(x\) and \(y\), the heat energy lost by liquid \(y\)

will be equal to the heat energy absorbed by liquid \(x\).

Consider the specific heat capacity of liquid \(\mathrm{x}\) is \(s_x\), of liquid \(\mathrm{y}\) is \(s_y\)

and of liquid \(\mathrm{z}\) is \(s_z\).

Now, for liquid \(\mathrm{x}\) and \(\mathrm{y}\), using principle of calorimetry,

Heat gained by liquid \(x=\) Heat lost by liquid \(y\)

\(m s_x\left(\mathrm{~T}_1-10\right)=m s_y\left(20-\mathrm{T}_1\right) \)

\(\Rightarrow s_x(16-10)=s_y(20-16) \)

\(\Rightarrow \frac{s_x}{s_y}=\frac{2}{3} \ldots(\mathrm{i})\)

Similarly, for liquid \(y\) and \(z\), using principle of calorimetry,

Heat gained by liquid \(\mathrm{y}=\) Heat lost by liquid \(\mathrm{z}\)

\(m s_y\left(\mathrm{~T}_2-20\right)=m s_z\left(30-\mathrm{T}_2\right) \)

\(\Rightarrow s_y(26-20)=s_z(30-26) \ldots \)

\(\Rightarrow \frac{s_y}{s_z}=\frac{2}{3} \ldots \text { (i) }\)

Multiply Eq. (i) by Eq. (ii), we get

\(\frac{s_x}{s_z}=\frac{4}{9}\)

Consider the mixing of liquids \(x\) and \(z\), let the temperature of mixture be \(T_3\).

Using principle of calorimetry for liquids \(x\) and \(z\).

Heat gained by liquid \(x=\) Heat lost by liquid \(z\)

\(m s_x\left(\mathrm{~T}_3-10\right)=m_z\left(30-\mathrm{T}_3\right)\)

\(\Rightarrow s_x\left(\mathrm{~T}_3-10\right)=s_z\left(30-\mathrm{T}_3\right)\)

\(\Rightarrow \frac{s_x}{s_z}=\frac{30-\mathrm{T}_3}{\mathrm{~T}_3-10} \)

\(\Rightarrow \frac{4}{9}=\frac{30-\mathrm{T}_3}{\mathrm{~T}_3-10} \)

\(\Rightarrow 4 \mathrm{~T}_3-40=270-9 \mathrm{~T}_3 \)

\(\Rightarrow \mathrm{T}_3=23.84^{\circ} \mathrm{C}\)

We know that:

The relation between the ratio of \(C_{p}\) and \(C_{V}\) with a degree of freedom is given by,

\(\gamma=\frac{C_{p}}{C_{v}}=1+\frac{2}{f}\)

Where, \(C_{v}=\) molar specific heat capacity of a gas at constant volume, \(C_{p}=\) molar specific heat of a gas at constant pressure, \(f=\) degree of freedom, \(\gamma\) is the adiabatic constant

Diatomic gas has 5 degrees of freedom,

\(\gamma=1+\frac{2}{5}\)

\(=\frac{5+2}{5}\)

\(=\frac{7}{5}\)

When a ferromagnetic material is heated to Curie temperature, it disrupts the arrangements of the molecules and a weak magnetic behavior remains.This weak magnetic behavior is called Paramagnetic.Above this temperature, the paramagnetism property also decreases.Upon cooling, it regains its ferromagnetic behavior.