Physics Test - 40

Coulomb’s law: When two charged particles of charges \(\mathrm{q}_{1}\) and \(\mathrm{q}_{2}\) are separated by a distance \(\mathrm{r}\) from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.

From the above, it is clear that Coulomb's force between two point charges varies with distance '\(\mathrm{r}\)' in relation to \(\mathrm{\frac{1}{r^{2}}}\).

lf the speed of rotation of the earth about its axis ln the weight of the body at the equator will decrease.

This is due to the gravity's force is resolved into a centripetal force that acts parallel to the equator.Thus, at the Poles, apparent weight is the same as \(mg\).

At the equator, apparent weight will be \(m g^{\prime}=\mathrm{m}\left(g-R \omega^{2}\right)\)

Where, \(\omega\) is the angular velocity.

As we know,

\(1 \mathrm{~kg}=1000 ~\mathrm{gm}\)

\( 1 \mathrm{~m}=100 \mathrm{~cm}\)

\( 1 \) Newton\(=1~ \mathrm{kg}~ \mathrm{m} / \mathrm{s}^{2}\)

\( 1 \) Newton \(=1000~ \mathrm{gm} \times 100 \mathrm{~cm} / \mathrm{s}^{2}\)

\( 1 \) Newton \( = 10^{5}~ \mathrm{gm}~\mathrm{cm} / \mathrm{s}^{2}\)

\( 1 ~ \mathrm{gm}~\mathrm{cm} / \mathrm{s}^{2} = 10^{-5}\) Newton

As we know,

\(1\) dyne \(=1~\mathrm{gm} ~ \mathrm{cm} / \mathrm{s}^{2}\)

Then,

\(1 \) Newton \(=10^{5}\) dyne

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings.Thus, the process is adiabatic.
As we know that, the value of the ratio of specific heat for standard gas\(=1.40\).
\(\therefore \gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\mathrm{1 . 4}\)
For an adiabatic process, we have:\(\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{V}_{2}^{\gamma}\)
The final volume is compressed to half of its initial volume.
\(\therefore \mathrm{V}_{2}=\mathrm{V}_{1} / \mathrm{2}\)
\(\mathrm{P}_{1} \mathrm{V}_{1}^{\gamma}=\mathrm{P}_{2}\left(\mathrm{V}_{1} / \mathbf{2}\right)^{\gamma}\)
\(\mathrm{P}_{2} / \mathrm{P}_{1}=\mathrm{V}_{1}^{\gamma} /\left(\mathrm{V}_{1} / \mathrm{2}\right)^{\gamma}\)
\(=\mathrm{2}^{\mathrm{1 . 4}}=\mathrm{2 . 6 3 9}\)

Let total distance be \(S\)

\(S\)= \(1350\) \(m\)

Let \(a=1 ~m / s^{2}\)

\(r=3 ~m/s^{2}\)

Let \(S_{1}\) be the distance covered with positive acceleration

Let \(S_{2}\) be the distance travelled with negative acceleration

We know \(v^{2}-u^{2}=2 ~a s\)

Therefore, \(S_{1}=\frac{\left(v^{2}-u^{2}\right)}{2a}\) ……..\((i)\)

And \(S_{2}=\frac{\left(v^{2}-u^{2}\right)}{2r}\) ……..\((ii)\)

Now, \(v\) and \(u\) are same for the train

There \(v^{2}-u^{2}\) is same for both equations

Equation \((i)\) divided by \((ii)\) give

\(\frac{S_{1}}{S_{2}}=\frac{r}{a}\)

i.e., \(\frac{S_{1}}{S_{2}}=\frac{3}{1}\)

\(S_{1}=3 ~S_{2}\)

Now, \(S_{2}=S-S_{1}\)

Therefore, \(S_{1}=3\left (S-S_{1}\right)\)

l.e, \(S_{1}=3 S-3 S_{1}\)

\(4 S_{1}=3 S\)

\(S_{1}=\frac{3}{4} × S = \frac{(3 × 1350)}{4} = 1012.5 ~m\)

And so \(S_{2}=\frac{1}{4} × S = \frac{1350}{4} = 337.5 ~m\)

Now, \(S_{1}=u+\frac{1}{2} × a × t^{2}\)

\(u=0\) as train starts from rest

\(1012.5=\frac{1}{2} × 1 × t_{1}^{2}\)

\(t_{1}^{2}=2025\)

\(t_{1}=45~s \)

Also \(S_{2}=\frac{1}{2} × r × t_{2}^{2}\)

\(337.5=\frac{3}{2} × t_{2}^{2}\)

\(t_{2}^{2}=225\)

\(t_{2}=15 ~s \)

Therefore, total time

\(T=t_{1}+t_{2}\)

\(T=45+15\)

\(T=60~s \)

Therefore, total time taken by train \(=60~s \)

Radius of air bubble; \(R =0.1 mm\)

\(=0.1 \times 10^{-3} m \left(1 mm =10^{-3} m \right)\)

Surface tension of water, \(T =7.2 \times 10^{-2} N \mid m\).

The excess pressure inside an air bubble is given by \(: \rightarrow\)

\(P_{2}-P_{1}=\frac{2 T} {R}\)

\(P_{2}=\) Pressure inside air bubble

\(P_{1}=\) Atmospheric pressure

\(P _{2}- P _{1}=\frac{2 \times\left(7.2 \times 10^{-2}\right)}{0.1 \times 10^{-3}}\)

\(=1.44 \times 10^{3} N / m ^{2}\)

Now, \(P _{2}= P _{1}+1.44 \times 10^{3} N / m ^{2}\)

\(=1.013 \times 10^{5}+1.44 \times 10^{3}\)

\(=1.027 \times 10^{5} N / m ^{2}\)

Total displacement \(=\left(v_{1} × \frac{t}{2} \right)+\left(v_{2} × \frac{t}{2}\right)\)

\(=\left(v_{1}+v_{2}\right) \frac{t}{2}\)

Mean velocity \(=\) \(\frac{\text { Total displacement }}{\text { Total time }}\)

\(=\frac{\left[\left(v_{1}+v_{2}\right) \frac{t}{2}\right]}{t}\)

\(=\frac{\left(v_{1}+v_{2}\right)}{2}\)

Mean velocity of man is \(\frac{\left(v_{1}+v_{2}\right)}{2}\)

Given, the length of the water pipe, L = 1 m
The cross-sectional area of the water pipe, A=1cm2=10−4m2
The temperature of the ice = -10ºC
Current passing in the conductor, I = 0.5 A
Resistance of the conductor, R = 4 kΩ
The latent heat of fusion for ice, ,Li=3.33×105J∕kg
The density of the ice, d =1000kg∕m3
The specific heat of the ice, cp,ice=2×103J∕kg
Heat required to melt the ice at 10°C to 0°C
Q=mcpΔT+mLi
Q=dVcpΔT+dVLf

(∵V=A×L)
= 35300 J
According to the Joule’s law of heating,
H=I2Rt
⇒ 35300=(0.5)2(4000)(t)
t = 35.3 s
Thus, the minimum time required to melt the ice is 35.3 s.

Let us assume that the threshold frequency of the sphere is \(\lambda_{o} .\) Let the stopping potential is \(V'\) of the surface when the light of wavelength \(\lambda_{3}\) is used. Thus according to Einstein photoelectric equation.

\(\frac{h c}{\lambda_{1}}=\frac{h c}{\lambda_{o}}+e V\).....(i)

\(\frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{o}}+3 e V\).....(ii)

\(\frac{h c}{\lambda_{3}}=\frac{h c}{\lambda_{o}}+e V^{\prime}\).....(iii)

From equation (i) and (ii)

\(\frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{0}}+3\left(\frac{h c}{\lambda_{1}}-\frac{h c}{\lambda_{0}}\right)\)

\(\Rightarrow \frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{0}}+\frac{3 h c}{\lambda_{1}}-\frac{3 h c}{\lambda_{0}}\)

\(\Rightarrow \frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-\frac{2 h c}{\lambda_{0}}\)\(\ldots(i v)\)

From equation (iii) and (iv)

\(\frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-2\left(\frac{h c}{\lambda_{3}}-e V^{\prime}\right)\)

\(\Rightarrow \frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-\frac{2 h c}{\lambda_{3}}+2 e V^{\prime}\)

\(\Rightarrow h c\left(\frac{1}{\lambda_{2}}+\frac{2}{\lambda_{3}}-\frac{3}{\lambda_{1}}\right)=2 e V^{\prime}\)

\(\Rightarrow \frac{h c}{e}\left(\frac{1}{ \lambda_{3}}+\frac{1}{2\lambda_{2}}-\frac{3}{2 \lambda_{1}}\right)=V^{\prime}\)

From the given graph, slope of F-x graph is k.

 \(k=\frac{F}{x}\)

\(=\frac{80}{0.2}\)

\(=400 \mathrm{Nm}^{-1}\)

\(m=0.01\)

We know that time period \(T\) is given by:

\(T=2 \pi \sqrt{\frac{m}{k}}\)

\(=2 \pi \sqrt{\frac{0.01}{400}}\)

\(=\frac{\pi}{100}\)

\(=0.03142 \mathrm{~s}\)