When two identical conducting spheres are in contact, their charges are shared equally. The total charge on both spheres is: \[ Q_{\text{total}} = 4 \times 10^{-6} \, \text{C}. \] Thus, the charge on each sphere after they are in contact will be: \[ Q = \frac{4 \times 10^{-6}}{2} = 2 \times 10^{-6} \, \text{C}. \] Using Coulomb's law for the force of repulsion between the two spheres: \[ F = \frac{1}{4\pi \epsilon_0} \frac{Q^2}{r^2}. \] Substitute the known values for the force and charge and solve for the distance \( r \): \[ 9 \times 10^{-3} = \frac{9 \times 10^9 \times (2 \times 10^{-6})^2}{r^2}. \] Solving for \( r \), we find \( r = 4 \, \text{cm} \).
Final Answer: \( 4 \, \text{cm} \).
A tube of length L is shown in the figure. The radius of cross section at point (1) is 2 cm and at the point (2) is 1 cm, respectively. If the velocity of water entering at point (1) is 2 m/s, then velocity of water leaving the point (2) will be:
In photoelectric effect, the stopping potential \( V_0 \) vs frequency \( \nu \) curve is plotted. \( h \) is the Planck's constant and \( \phi_0 \) is the work function of metal.
(A) \( V_0 \) vs \( \nu \) is linear.
(B) The slope of \( V_0 \) vs \( \nu \) curve is \( \frac{\phi_0}{h} \).
(C) \( h \) constant is related to the slope of \( V_0 \) vs \( \nu \) line.
(D) The value of electric charge of electron is not required to determine \( h \) using the \( V_0 \) vs \( \nu \) curve.
(E) The work function can be estimated without knowing the value of \( h \).
Choose the correct answer from the options given below:
Match List - I with List - II:
List - I:
(A) Amylase
(B) Cellulose
(C) Glycogen
(D) Amylopectin
List - II:
(I) β-C1-C4 plant
(II) α-C1-C4 animal
(III) α-C1-C4 α-C1-C6 plant
(IV) α-C1-C4 plant
Match List - I with List - II:
Choose the correct answer from the given below options