Physics Test - 71

Since here total flux entering equals the total flux leaving and hence net flux equals zero.

Actual weight on the equator \(=W=mg\)

where \('m'\) is the mass and \('g'\) is the gravity.

According to the given condition,

Weight on the equator \(=W^{\prime}=mg^{\prime}\)

Weight on the equator \(=W^{\prime}=\frac{3}{5} mg \ldots \ldots \ldots .\) \((i)\)

We know that, \(\lambda=0\) at the equator.

\(mg^{\prime}=mg-m R \omega^{2} \cos \lambda\)

\(\Rightarrow \frac{3}{5} mg=mg-m R \omega^{2} \cos 0(\because \lambda=0)\)

\(\Rightarrow \frac{3}{5} mg=mg-m R \omega^{2}(1)(\because \cos 0=1)\)

\(\Rightarrow \frac{3}{5} mg=mg-m R \omega^{2}\)

\(\Rightarrow m R \omega^{2}=mg-\frac{3}{5} mg\)

\(\Rightarrow m R \omega^{2}=(1-\frac{3}{5}) mg\)

\(\Rightarrow mR \omega^{2}=\frac{2}{5} mg\)

\(\Rightarrow R \omega^{2}=\frac{2}{5} g\)

\(\Rightarrow \omega^{2}=\frac{2 g}{5 R}\)

\(\Rightarrow \omega^{2}=\frac{(2 × 10)}{\left(5 × 6.4 × 10^{5}\right)}\)

\( \because R=\) Radius of Earth \(=6400 ~km=\)\(\left.6.4 × 10^{6} ~m\right)\)

\(\Rightarrow \omega=\sqrt{\left(6.25 × 10^{-7}\right)}\)

\(\Rightarrow \omega=7.8 × 10^{-4} ~rad / sec\)

We know that \(Y=3 K(1-2 \sigma)\)

Here,

\(\sigma=\) Poisson's ratio

\(\sigma=\frac{1}{2}\left(1-\frac{Y}{3 K}\right)\)

Also,

\(Y=2 \eta(1+\sigma) \)

\( \sigma=\frac{Y}{2 \eta}-1\)

From equation (i) and (ii), we have

\( \frac{1}{2}\left(1-\frac{Y}{3 K}\right)=\frac{Y}{2 \eta}-1 \)

\( \Rightarrow 1-\frac{Y}{3 K}=\frac{Y}{\eta}-2 \Rightarrow \frac{Y}{3 K}=3-\frac{Y}{\eta} \)

\( \Rightarrow \frac{Y}{\eta-Y} \)

\( \Rightarrow \frac{\eta Y}{3 K}=3 \eta-Y\)

\(\Rightarrow K=\frac{\eta Y}{9 \eta-3 Y}\)

\(\text{R}_1+\text{R}_2= 1000\)

\(\Rightarrow \text{R}_2 = 1000 - \text{R}_1\)

On balancing condition

\(\text{R}_1 (100-1) = (1000 - \text{R}_1)1 ......\text{(i)}\)

On Interchanging resistance balance point shifts left by 10cm

On balancing condition

\(\left(1000-R_1\right)(110-1)=R_1(1-10) \)

\(\text { or, } R_1(1-10)=\left(1000-R_1\right)(110-1).......\text{(ii)}\)

Dividing eqn (i) by (ii)

\(\frac{100-1}{1-10}=\frac{1}{110-1} \)

\(\Rightarrow(100-1)(110-1)=1(1-10) \)

\(\Rightarrow 11000-1001-1101+1^2=1^2-101 \)

\(\Rightarrow 11000=2001\)

or, \(1=55\)

Putting the value of ' 1 ' in eqn (i)

\(\mathrm{R}_1(100-55)=\left(1000-\mathrm{R}_1\right) 55 \)

\(\Rightarrow \mathrm{R}_1(45)=\left(1000-\mathrm{R}_1\right) 55 \)

\(\Rightarrow \mathrm{R}_1(9)=\left(1000-\mathrm{R}_1\right) 11 \)

\(\Rightarrow 20 \mathrm{R}_1=11000 \)

\(\therefore \mathrm{R}_1=550 \mathrm \Omega\)

When the widths of the two slits are increased, the fringes become brighter. However, the width of each slit should be considerably smaller than the separation between the slits. When the slits become so wide that this condition is not satisfied, the interference pattern disappears i.e., The fringes become less distinct.

A laser device produces amplification in the Ultraviolet or visible region.

Laser, a device that stimulates atoms or molecules to emit light at particular wavelengths and amplifies that light, typically producing a very narrow beam of radiation. The emission generally covers an extremely limited range of visible, infrared, or ultraviolet wavelengths.

The \(P-V\) diagram of adiabatic process, isothermal process and isobaric process. 

Work done in process= area enclosed by \(P-V\) diagram with volume axis.Since area under the curve is maximum for adiabatic process, so work done on the gas will be maximum for adiabatic process.

The universal constant of gravity is not one that depends on the nature of the medium in which the bodies are placed.

The gravitational constant can be defined as the constant relating the force exerted on the objects to the mass and distance between the objects. The value of the universal gravitation constant is found to be \(\mathrm{G}=6.673 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\).