An oil drop carrying a charge of 2 electrons has a mass of 3.2 × 10^-17 kg. It is falling freely in air with terminal speed…

Q: An oil drop carrying a charge of 2 electrons has a mass of 3.2 × 10-17 kg. It is falling freely in air with terminal speed. The electric field required to make the drop move upwards with the same speed is

(a) 2 × 103 V/m

(b) 4 × 103 V/m

(c) 3 × 103 V/m

(d) 8 × 103 V/m

Ans:(a)

Sol: Without electric field when oil drop is moving downwards,

mg = 6 π η r v …(i)

Under electric field of strength E , oil drop is to move upwards

F – mg = 6 π η r v

F – mg = mg …using (i)

(2 e)E = 2 mg

E = mg/e

\displaystyle E = \frac{3.2 \times 10^{-17}\times 10}{1.6 \times 10^{-19}}

= 2 × 103 V /m

A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected…

Q: A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q , E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done by the system in question, in the process of inserting the slab, then select the incorrect relation from the following:

(a) \displaystyle Q = \frac{\epsilon_0 A V}{d}

(b) \displaystyle E = \frac{ V}{K d}

(c) \displaystyle W = \frac{\epsilon_0 A V^2}{2 K d}

(d) \displaystyle W = \frac{\epsilon_0 A V^2}{2 d}(1-\frac{1}{K})

Ans: (c)

Sol: As battery is disconnected; Q remains the same

\displaystyle Q = C_0 V = \frac{\epsilon_0 A V}{d}

Electric intensity,\displaystyle E = \frac{V'}{d} = \frac{V}{K d}

Work done = U1 – U2

\displaystyle  = \frac{1}{2}\frac{\epsilon_0 A}{d} V^2 - \frac{1}{2}\frac{K \epsilon_0 A}{d} \frac{V^2}{K^2}

\displaystyle  = \frac{\epsilon_0 A V^2}{2 d} (1-\frac{1}{K})

A 100 eV electron is fired directly towards a large metal plate having surface charge density…

Q: A 100 eV electron is fired directly towards a large metal plate having surface charge density -2 × 10-6 cm-2. The distance from where the electron be projected so that it just fails to strike the plate is

(a) 0.22 mm

(b) 0.44 mm

(c) 0.66 mm

(d) 0.88 mm

Ans:(b)

Sol: Here, K.E. = 100 eV = 100 × 1.6 ×10-19 joule. This is lost when electron moves through a distance d towards the negative plate.

K.E. = work done = qE × s

\displaystyle K.E = e (\frac{\sigma}{\epsilon_0}) d ; Where σ = surface charge dinsity .

\displaystyle d = \frac{K.E \times \epsilon_0}{e \times \sigma}

\displaystyle d = \frac{100\times 1.6 \times 10^{-19}\times 8.85 \times 10^{-12}}{1.6 \times 10^{-19} \times 2 \times 10^{-6}}

= 4.43 × 10-4 m

= 0.44 mm

The order of energies of energy levels, A, B and C is EA < EB < EC , if the wavelength corresponding to transitions...

Q: The order of energies of energy levels, A, B and C is EA < EB < EC , if the wavelength corresponding to transitions C → B, B → A and C → A are λ1 , λ2 and λ3 respectively , then which of the following relations is correct ?

(a) λ1 + λ2 + λ3 = 0

(b) λ32 = λ12 + λ22

(c) λ3 = λ1 + λ2

(d) \displaystyle \lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}

Ans: (d)

Sol: ECA = ECB + EBA

\displaystyle \frac{h c}{\lambda_3} = \frac{h c}{\lambda_1} + \frac{h c}{\lambda_2}

\displaystyle \frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}

\displaystyle \lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}

A deutron of atomic mass 2·0147 a.m.u. and negligible kinetic energy is absorbed by a Li6 nucleus…

Q: A deutron of atomic mass 2·0147 a.m.u. and negligible kinetic energy is absorbed by a Li6 nucleus of mass 6·0169 a.m.u., the intermediate nucleus disintegrates spontaneously into two α – particles, each of mass 4·0039 a.m.u. The energy transferred to each α particle is

(a) 12·08 MeV

(b) 11.08 MeV

(c) 6·04 MeV

(d) 5·54 MeV

Ans: (b)

Sol: The given nuclear reaction is

3Li6 + 1H2  → 2 (2He4 )

Mass defect, Δm = (6.0169+2.0147)-2×4.0039

=(8.0316-8.0078) amu

Δm = 0.0238 amu

Energy available = 0.0238 × 931 MeV

KE of each α particle =(0.0238×931)/2 MeV

= 11.08 MeV