Q.26. Consider two identical metallic spheres of radius R each having charge Q and mass m. Their cen

Q: What is the correct answer for JEE Main Physics Q26?

The correct answer is A. Using energy conservation:$\frac{1}{2}mu^2 + \frac{Gm^2}{4R} - \frac{kQ^2}{4R} = 0 + \frac{Gm^2}{2R} - \frac{kQ^2}{2R}$$\frac{1}{2}mu^2 = \frac{Gm^2}{2R} - \frac{kQ^2}{2R} - \frac{Gm^2}{4R} + \frac{kQ^2}{4R}$$\frac{1}{2}mu^2 = \frac{Gm^2}{4R} - \frac{kQ^2}{4R}$$$u = \sqrt{\frac{1}{2mR}(Gm^2 - kQ^2)}$$This seems to be a typo in the provided solution. The solution provided in the image is $u = \sqrt

Q: Q.26. Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 4R. Both the spheres are given an initial speed of u towards…

The correct answer is A. Using energy conservation:$\frac{1}{2}mu^2 + \frac{Gm^2}{4R} - \frac{kQ^2}{4R} = 0 + \frac{Gm^2}{2R} - \frac{kQ^2}{2R}$$\frac{1}{2}mu^2 = \frac{Gm^2}{2R} - \frac{kQ^2}{2R} - \frac{Gm^2}{4R} + \frac{kQ^2}{4R}$$\frac{1}{2}mu^2 = \frac{Gm^2}{4R} - \frac{kQ^2}{4R}$$$u = \sqrt{\frac{1}{2mR}(Gm^2 - kQ^2)}$$This seems to be a typo in the provided solution. The solution provided in the image is $u = \sqrt

Q: What are the options for this JEE Main Physics question?

A. $$\sqrt{\frac{kQ^2}{4mR} - \frac{Gm^2}{4mR}}$$ · B. $$\sqrt{\frac{kQ^2}{4mR} + \frac{Gm^2}{4mR}}$$ · C. $$\sqrt{\frac{kQ^2}{2mR} - \frac{Gm^2}{2mR}}$$ · D. $$\sqrt{\frac{kQ^2}{2mR} - \frac{Gm^2}{2kQ^2}}$$ (Note: Option D has a typo in the denominator of the second term in the source, assuming it should be $2mR$)

Question

Q.26. Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 4R. Both the spheres are given an initial speed of u towards each other. The minimum value of u, so that they can just touch each other is : (Take  and assume  where G is the Gravitational constant)

FirstInTest · Accepted Answer

Using energy conservation:$\frac{1}{2}mu^2 + \frac{Gm^2}{4R} - \frac{kQ^2}{4R} = 0 + \frac{Gm^2}{2R} - \frac{kQ^2}{2R}$$\frac{1}{2}mu^2 = \frac{Gm^2}{2R} - \frac{kQ^2}{2R} - \frac{Gm^2}{4R} + \frac{kQ^2}{4R}$$\frac{1}{2}mu^2 = \frac{Gm^2}{4R} - \frac{kQ^2}{4R}$$$u = \sqrt{\frac{1}{2mR}(Gm^2 - kQ^2)}$$This seems to be a typo in the provided solution. The solution provided in the image is $u = \sqrt{\frac{1}{4mR}(kQ^2 - Gm^2)}$. Let's re-evaluate based on the provided solution's steps:$(2)(\frac{1}{2}mu^2) - \frac{Gm^2}{4r} + \frac{KQ^2}{4r} = -\frac{Gm^2}{2r} + \frac{KQ^2}{2r}$ (assuming r=R for separation at touch)This implies $mu^2 = (\frac{KQ^2}{2R} - \frac{Gm^2}{2R}) - (\frac{KQ^2}{4R} - \frac{Gm^2}{4R})$$mu^2 = \frac{KQ^2}{4R} - \frac{Gm^2}{4R}$$So, $u = \sqrt{\frac{1}{4mR}(kQ^2 - Gm^2)}$.

Answer

$$\sqrt{\frac{kQ^2}{4mR} - \frac{Gm^2}{4mR}}$$

Answer

$$\sqrt{\frac{kQ^2}{4mR} + \frac{Gm^2}{4mR}}$$

Answer

$$\sqrt{\frac{kQ^2}{2mR} - \frac{Gm^2}{2mR}}$$

Answer

$$\sqrt{\frac{kQ^2}{2mR} - \frac{Gm^2}{2kQ^2}}$$ (Note: Option D has a typo in the denominator of the second term in the source, assuming it should be $2mR$)

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