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physics ncert exemplar solutions class 12th chapter two

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Electrostatic Potential and Capacitance

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will:

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alok kumar singh

Contributor-Level 10

UI= $\frac{1}{2} c v^{2}$

$U_{i} = \frac{1}{2} \frac{\in_{0} A}{d} * k v^{2}$

$U_{i} = \frac{1}{2} \frac{\in_{0} A}{d} * 10 v^{2}$

$U_{f} = \frac{1}{2} \frac{\in_{0} A}{d} * 15 v^{2}$

$\frac{U_{f}}{U_{i}} = \frac{3}{2}$

$\frac{U_{f} - U_{i}}{U_{i}} * 100 = (\frac{3}{4} - 1) * 100$

$\frac{1}{2} * 100$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Electrostatic Potential and Capacitance

A composite parallel plate capacitor is made up to two different dielectric materials with different thickness (t₁ and t₂) as shown in figure. The two different dielectric materials are separated by a conducting foil F. The voltage of the conducing foil is ___________V.

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Vishal Baghel

Contributor-Level 10

Capacitance of each capacitor

$C_{1} = \frac{A 3 ε_{0}}{\frac{1}{2}} = 6 A ε_{0}$

$C_{2} = A 4 ε_{0} = 4 A ε_{0}$

Equivalent capacitance

$Δ v_{2} = \frac{240 A ε_{0}}{4 A ε_{0}} = 60 v$

$v_{f o i l} = 60 v$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Electrostatic Potential and Capacitance

A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key 'K' is closed, the total energy stored across the combination is E₁. Now key 'K' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E₂. The ratio E₁/E₂ will be:

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alok kumar singh

Contributor-Level 10

I > When switch is closed

Ceq = 2C

Energy E1 = $\frac{1}{2} * C_{e q} * v^{2} = \frac{1}{2} * 2 C * v^{2}$

E1 = CV2

II > When switch is opened

$E_{2} = \frac{1}{2} * (5 c) * V^{2} + \frac{{(C v)}^{2}}{2 * 5 C}$

$= \frac{13}{5} C V^{2}$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{5}{13}$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Units and Measurement

A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5mm and circular scale reading is 7. The calculated curved surface area wire to appropriate significant figures is:

[Screw gauge has 50 divisions on its circular scale]

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Vishal Baghel

Contributor-Level 10

Least count = $\frac{0.5}{50} m m$ = 0.01 mm

Diameter, d = 1.5 + 7 Ã— 0.01

= 1.57

$∴$ Surface Area = (2r) $l$

= 3.4 cm²

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2 months ago

physics ncert exemplar solutions class 12th chapter two Electrostatic Potential and Capacitance

27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is……… times that of a smaller drop.

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Payal Gupta

Contributor-Level 10

With the help of conservation of volume, we can write

$27 * \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 3 r . . . . . . . (1)$

With the help of conservation of charge, we can write

Q = 27 q. (2)

Potential energy of single drop = U1 = $\frac{q^{2}}{8 π ε_{0} r}$

Potential energy of bigger drop = $U_{2} = \frac{Q^{2}}{8 π ε_{0} R} = \frac{27 * 27 * q^{2}}{8 π ε_{0} (3 r)} = 243 (\frac{q^{2}}{8 π ε_{0} r}) = 243 U_{1}$

$\Rightarrow \frac{U_{2}}{U_{1}} = 243$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Electrostatic Potential and Capacitance

The total charge on the system of capacitors $C_{1} = 1 μ F, C_{2} = 2 μ F, C_{3} = 4 μ F a n d C_{4} = 3 μ F$ connected in parallel is:

(Assume a battery of 20V is connected to the combination)

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Vishal Baghel

Contributor-Level 10

$C_{e q} = C_{1} + C_{2} + C_{3} + C_{4}$

= 1 + 2 + 4 + 3

$= 10 ? F$

Q = C_eq v

= (10 x 20) $? C$

Q = 200 $? C$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Electrostatic Potential and Capacitance

Two uniformly charged spherical conductors A and B of radii 5mm and 10mm are separated by a distance of 2cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere A and B will be:

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alok kumar singh

Contributor-Level 10

$V_{A} = V_{B}$

$\Rightarrow \frac{k Q_{A}}{R_{A}} = \frac{k Q_{B}}{R_{B}}$

$\Rightarrow \frac{Q_{A}}{Q_{B}} = \frac{R_{A}}{R_{B}} * {(\frac{R_{B}}{R_{A}})}^{2} = \frac{1}{2} * {(\frac{2}{1})}^{2} = \frac{2}{1}$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Electrostatic Potential and Capacitance

If 'C' and 'V' represent capacity and voltage respectively then what are the dimensions of $λ$ where C/V = $λ ?$

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Payal Gupta

Contributor-Level 10

$λ = \frac{c}{v} = \frac{Q^{3}}{W^{2}} = \frac{I^{3} T^{3}}{M^{2} L^{4} T^{- 4}} \Rightarrow [λ] = [M^{- 2} L^{- 4} T^{7} l^{3}]$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Units and Measurement

In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. Then both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4^th Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6^th Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical bo

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alok kumar singh

Contributor-Level 10

1 MSD = 1mm

9 MSD = 10 VSD

1VSD = 0.9 MSD = 0.9 mm

Least count = 1 MSD – 1 VSD

= 0.1 mm = 0.01 cm

Zero Error = (10 – 4) * 0.1 = 0.6 mm

Reading = MSR + VSR – Zero error

= 3 cm + 6 * 0.01 – (0.06)

= 3.12 cm $?$ 3.10 cm

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2 months ago

physics ncert exemplar solutions class 12th chapter two Electrostatic Potential

A two point charges $4 q$ and $- q$ are fixed on the $x$ -axis at $x = - \frac{d}{2}$ and $x = \frac{d}{2}$ , respectively. If a third point charge ' $q$ ' is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will:

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alok kumar singh

Contributor-Level 10

$U_{initial} = \frac{k (4 q) (q)}{(d / 2)} + \frac{k (q) (- q)}{(d / 2)}$

$\begin{array}{r} \Rightarrow \frac{6 k q^{2}}{d} \\ \Rightarrow U_{final} = \frac{4 (4 q) (q)}{(\frac{3 d}{2})} + \frac{k (q) (- q)}{(d / 2)} \\ \Rightarrow \frac{2}{3} \frac{k q^{2}}{d} \\ \Rightarrow Δ U = (\frac{2}{3} - 6) \frac{k q^{2}}{d} \Rightarrow \frac{- 16}{3} \frac{{k q}^{2}}{d} \end{array}$

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