This problem involves converting a galvanometer into an ammeter using a shunt resistor. We need to calculate the value of this shunt resistor.
To convert a galvanometer into an ammeter, a low resistance shunt ($S$) is connected in parallel with the galvanometer. The formula to calculate the shunt resistance is:
$S = \frac{G \times I_g}{I - I_g}$
Substitute the given values into the formula:
$S = \frac{100 \ \Omega \times (1 \times 10^{-3} \text{ A})}{10 \text{ A} - (1 \times 10^{-3} \text{ A})}$
Simplify the numerator:
$100 \ \Omega \times 0.001 \text{ A} = 0.1 \text{ V}$
Simplify the denominator:
$10 \text{ A} - 0.001 \text{ A} = 9.999 \text{ A}$
Calculate the shunt resistance $S$:
$S = \frac{0.1 \text{ V}}{9.999 \text{ A}} \approx 0.010001 \ \Omega$
Rounding to a practical value, the required shunt resistance is approximately $0.01 \ \Omega$.
The calculated shunt resistance required to convert the galvanometer into an ammeter with a range of 0-10 A is $0.01 \ \Omega$.
In the circuit shown below, the voltage appearing across the diode D will be of the form :
A 100-turn closely wound circular coil of radius $10 \text{ cm}$ has a magnetic field of $3.14 \times 10^{-3} \text{ T}$ at its centre. The current passing through the coil, and the magnitude of the magnetic moment of this coil are, respectively :
(Take $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$)
The current $I$ in the circuit shown below is :
(All diodes are ideal and identical.

The figure given below shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current I. The current I is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field (B) with distance (r) from the axis of the conductor in the region is :
Five capacitors of capacitances $C_1 = C_2 = C_3 = C_4 = 10 \ \mu\text{F}$ and $C_5 = 2.5 \ \mu\text{F}$ are connected as shown along with a battery of $50 \text{ V}$. 
The equivalent capacitance and the charges on each capacitor respectively are :
Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by $15 \text{ cm}$ length of wire $Q$ is________.
$(\mu_o = 4\pi \times 10^{-7} \text{ T.m/A})$