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Step-by-Step Detailed Solution
Step 1: Identify the Forces in Air
When two identical charged spheres hang by identical strings making an angle of 37° with each other, each sphere experiences the following forces:
Gravitational force ( mg ) acting vertically downward.
Tension ( T ) in the string acting along the string’s length from the point of suspension.
Electrostatic repulsion ( F_e ) acting horizontally along the line connecting the two spheres.
Because the strings make a total angle of 37° with each other, each string is inclined at half of 37°, i.e., 37^\circ/2 = 18.5^\circ from the vertical (in symmetrical arrangement). However, we can treat all using general \theta for simplicity.
Step 2: Force Balance in Air
Let \theta be the angle each string makes with the vertical. Then the total angle between the two strings is 2\theta = 37^\circ . For one sphere:
Vertical force balance:
T \cos\theta = mg.
Horizontal force balance:
T \sin\theta = F_e.
Dividing the horizontal balance by the vertical balance,
\tan\theta = \frac{F_e}{mg}.
Hence,
F_e = mg \,\tan\theta.
Step 3: Forces in the Liquid
When the spheres are submerged in a liquid of density \rho_L = 0.7 \,\text{g/cm}^3 , they experience an upward buoyant force F_b given by
F_b = \rho_L \, V \, g,
where V is the volume of each sphere. Therefore, the effective weight of each sphere in the liquid is
mg - F_b = mg - \rho_L \, V \, g.
If the dielectric constant of the liquid is k , the electrostatic force between the charges in the liquid becomes
F_e' = \frac{F_e}{k}.
Step 4: Force Balance in the Liquid
Let T' be the tension in each string in the liquid. The angle \theta with the vertical stays the same. For one sphere:
Vertical balance in liquid:
T' \cos\theta = mg - \rho_L \, V \, g.
Horizontal balance in liquid:
T' \sin\theta = \frac{F_e}{k}.
Thus,
\tan\theta = \frac{\frac{F_e}{k}}{mg - \rho_L \, V \, g}.
Step 5: Equate Expressions for \tan\theta
From the first case (in air),
\tan\theta = \frac{F_e}{mg},
and from the second case (in liquid),
\tan\theta = \frac{F_e/k}{\,mg - \rho_L\, V \,g\,}.
Since they describe the same angle,
\frac{F_e}{mg} = \frac{\frac{F_e}{k}}{mg - \rho_L \, V \, g}.
We can cancel F_e on both sides, leading to
\frac{1}{mg} = \frac{1/k}{mg - \rho_L \, V \, g}.
Rearranging,
mg - \rho_L \, V \, g = \frac{mg}{k}.
Step 6: Relate Density and Volume
If the density of the spheres is \rho_B = 1.4 \,\text{g/cm}^3 , then
m = \rho_B \, V.
Substitute m = \rho_B \, V into the previous equation:
\rho_B \, V \, g - \rho_L \, V \, g = \frac{\rho_B \, V \, g}{k}.
Divide out the common factor V \, g :
\rho_B - \rho_L = \frac{\rho_B}{k}.
Hence, the dielectric constant k is
k = \frac{\rho_B}{\rho_B - \rho_L}.
Step 7: Substitute the Given Numerical Values
Here, \rho_B = 1.4 \,\text{g/cm}^3 and \rho_L = 0.7 \,\text{g/cm}^3 . Substituting,
k = \frac{1.4}{1.4 - 0.7} = \frac{1.4}{0.7} = 2.
Final Answer
The required dielectric constant of the liquid is 2 .
