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Physics NCERT Exemplar Solutions Class 11th Chapter Eight

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Class 12th

Match List I with List II

List II List II

A. Facsimile I. Static Document Image

B. Guided media Channel II. Local Broadcast Radio

C. Frequency Modulation

...more

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

Based on theory

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Class 12th

Identify the logic operation performed by the given circuit:

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alok kumar singh

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Physics Nuclei

Which of the following figure represents the variation of $I_{n} (\frac{R}{R_{0}})$ with I_nA (if R = radius of a nucleus and A = its mass number)

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alok kumar singh

Contributor-Level 10

We know, R = R_oA₃₌

_{$\Rightarrow \frac{R}{R_{O}} = A^{3}$}

Taking log both side

$l o g_{e} (\frac{R}{R_{O}}) = 3 l o g A$

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Dual Nature of Radiation and Matter

A proton, a neutron, an electron and an a-particle have same energy. If $λ_{p}, λ_{n}, λ_{e} a n d λ_{α}$ and the de Broglie's wavelengths of proton, neutron, electron and a particle respectively, then choose the correct relation form the following:

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alok kumar singh

Contributor-Level 10

$λ = \frac{h}{m v} = \frac{h}{\sqrt[]{2 m k . E}}$

$λ α \frac{1}{\sqrt[]{m}}$

$m_{α} > m_{n} > m_{p} > m_{e}$

$λ_{α} < λ_{n} < λ_{p} < λ_{e}$

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Work, Energy and Power

A pendulum of length 2m consists of a wooden bob of mass 50g. A bullet of mass 75g is fired towards the stationary bob with a speed $υ .$ The bullet emerges out of the bob with a speed $\frac{υ}{3}$ and the bob just completes the vertical circle. The value of $υ$ is_________ms^-¹.

(if g = 10 m/s²).

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Vishal Baghel

Contributor-Level 10

Com

m1v + m2 * 0 = m1v1 + m2v2

75v = $45 \frac{v}{3} + 50 * 10$ $\frac{}{}$

$\Rightarrow 75 * \frac{2 v}{3} = 50 * 10$

$\Rightarrow v = \frac{3 * 50 * 10}{75 * 2} = 10 m / s e c$

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Mechanical Properties of Fluids

The area of cross-section of a large tank is 0.5 m². It has a narrow opening near the bottom having area of cross-section 1 cm². A load of 25 kg is applied on the water, coming out of the opening at the time when the height of water level in the tank is 40cm above the bottom, will be_________cms^-¹. $[T a k e g = 10 m s^{- 2}]$

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Vishal Baghel

Contributor-Level 10

Bernoulli's equation between (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{z g} + z_{2}$

$\frac{P_{a t m} + m g / A}{ρ g} + 0 + 40 + 1 0^{- 2}$

$\frac{P_{a t m}}{ρ g} + \frac{v_{2}^{2}}{2 g} + 0 [v_{1} \approx 0]$

$\frac{m g}{A ρ g} + 40 * 1 0^{- 2} = \frac{v_{2}^{2}}{2 g}$

$\frac{m}{A ρ} + 40 * 1 0^{- 2} = \frac{v_{2}^{2}}{2 g}$

$\Rightarrow \frac{25}{0.5 * 1 0^{3}} + 40 * 1 0^{2} = \frac{V_{2}^{2}}{2 * 10}$

$\Rightarrow V_{2}^{2} = (5 * 1 0^{- 2} + 40 * 1 0^{- 2}) 20$

$V_{2}^{2} = 45 * 1 0^{- 2} * 20$

$V_{2}^{2} = 90 * 1 0^{- 1}$

$V_{2} = 3 m / s e c$

$= 300 c m / s e c$

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Communication Systems

The height of a transmitting antenna at the top of a tower is 25m and that of receiving antenna is, 49m. The maximum distance between them, for satisfactory communication is LOS (Line-Of-Sight) is $K \sqrt[]{5} * 1 0^{2} m .$ The value of K is___________.

(Assume radius of Earth is 64 * 10⁺⁵m) [Calculate upto nearest integer value]

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Vishal Baghel

Contributor-Level 10

Given, hT = 25m, R = 6400 km

hR = 49m

dm (Maximum distance for satisfactory communication)

dm = $\sqrt[]{2 R h_{T}} + \sqrt[]{2 R h_{R}}$ $_{}_{}$

dm = $\sqrt[]{2 * 6400 * 1 0^{3} * 25} + \sqrt[]{2 * 6400 * 1 0^{3} * 49}$ $^{}^{}$

= $\sqrt[]{5 * 1 0^{4} * 6400} + \sqrt[]{6400 * 1 0^{3} * 49 * 2}$ $^{}^{}$

= $192 \sqrt[]{5} * 1 0^{2}$ $^{}$

= $k \sqrt[]{5} * 1 0^{2}$ $^{}$

k= 192

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Class 12th

A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is _________nJ.

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Vishal Baghel

Contributor-Level 10

Q = CV

= 50 * 10^-12 * 100

= 5 * 10^-12 * 10³

Q = 5 * 10^-9C

U_i (Initial Energy)

= $\frac{Q^{2}}{2 C} = \frac{25 * 1 0^{- 18}}{2 * 50 * 1 0^{- 12}}$

= $\frac{1}{4} * 1 0^{- 6}$

Now capacitor is connected to an identical uncharged capacitor

kvL

$\frac{Q_{1}}{C} = \frac{Q_{2}}{C} = 0$

q₁ = q₂

Initial change Q₁ + Q₂ = Q

$\Rightarrow 2 Q_{1} = Q$

$Q_{1} = \frac{Q}{2}$

$u_{F} = \frac{Q_{1}^{2}}{2 C} + \frac{Q_{1}^{2}}{2 C}$

$= \frac{{(\frac{Q}{2})}^{2}}{2 C} + \frac{{(Q / 2)}^{2}}{2 C}$

$= 2 * \frac{Q^{2}}{4 * 2 C}$

...more

Q = CV

= 50 * 10^-12 * 100

= 5 * 10^-12 * 10³

Q = 5 * 10^-9C

U_i (Initial Energy)

= $\frac{Q^{2}}{2 C} = \frac{25 * 1 0^{- 18}}{2 * 50 * 1 0^{- 12}}$

= $\frac{1}{4} * 1 0^{- 6}$

Now capacitor is connected to an identical uncharged capacitor

kvL

$\frac{Q_{1}}{C} = \frac{Q_{2}}{C} = 0$

q₁ = q₂

Initial change Q₁ + Q₂ = Q

$\Rightarrow 2 Q_{1} = Q$

$Q_{1} = \frac{Q}{2}$

$u_{F} = \frac{Q_{1}^{2}}{2 C} + \frac{Q_{1}^{2}}{2 C}$

$= \frac{{(\frac{Q}{2})}^{2}}{2 C} + \frac{{(Q / 2)}^{2}}{2 C}$

$= 2 * \frac{Q^{2}}{4 * 2 C}$

$U_{f} = \frac{Q_{2}}{4 * 2 C}$

Loss in energy = U_i = U_t

= $\frac{Q^{2}}{2 C} - \frac{Q^{2}}{4 C}$

= $\frac{Q^{2}}{4 C}$

= $\frac{1}{2} [\frac{Q^{2}}{2 C}]$

$\frac{1}{2} * 0.25 * 1 0^{- 6}$

= 125 * 10^-9

= 125 nJ

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Class 12th

The current density in a cylindrical wire of radius 4mm is 4 * 10⁶Am^-². The current through the outer portion of the wire between radial distances $\frac{R}{2}$ and R is _________πA.

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Vishal Baghel

Contributor-Level 10

J (current density) = 4 * 10⁶ Am^-2

Area between radial distance $\frac{R}{2} t o R$

A = $π [R^{2} - \frac{R^{2}}{4}] = \frac{3 R^{2}}{4} π$

I = AJ

= $\frac{3 R^{2}}{4} π * 4 * 1 0^{6}$

= 3R² * 10⁶ πAmp

= 3 (4 * 10^-3)² * 10⁶ πAmp

= 3 * 16 * 10^-6 * 10⁶π Amp

= 48πA

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Physics NCERT Exemplar Solutions Class 11th Chapter Eight Class 12th

A cell, shunted by a8 $Ω$ resistance, is balanced across a potentiometer wire of length 3m. The balancing length is 2m when cell is shunted by 4 $Ω$ resistance. The value of internal resistance of the cell will be__________ $Ω$ .

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Vishal Baghel

Contributor-Level 10

R shunt Resistance

Given

CD = 3m

CF = 2m

Current through potentiometer wire

I = $\frac{v}{R_{0}} = \frac{V A}{ρ L}$ $\frac{}{_{}} \frac{}{}$ (1)

A Area of potentiometer wire

$ρ \to$ Resistivity of the wire

L Length of the potentiometer wire

Case 1 When 8 $Ω$ shunt is bal aced at 3m length

$V_{C D} = I R_{C D} = \frac{V}{ρ L} A * \frac{ρ C F}{A}$

$V_{C D} = \frac{V l}{L} = \frac{V * 3}{L}$

VCD = VAB = E – I1r

= $- \frac{E}{R + r} r = \frac{V}{L} * 3$ $\frac{}{} \frac{}{}$

$\Rightarrow E [1 - \frac{r}{8 + r}] = \frac{V}{L} * 3$ (2)

Case 2 When 4 $Ω$ shunt is balanced at 2m length

$V_{C F} = I R_{C F} = \frac{V}{ρ L} A * \frac{ρ C F}{A} = \frac{V * 2}{L}$

$V_{C F} = V_{A B} = E - I_{2} r$

= $E - \frac{E}{R + r} r$ $\frac{}{}$

= $E [1 - \frac{r}{4 + r}]$ $\frac{}{}$

$\frac{}{} \frac{}{}$ $\Rightarrow E (1 - \frac{r}{4 + r}) = \frac{V * 2}{L}$ (3)

$(2) \div (3)$

$\frac{1 - \frac{r}{8 + r}}{1 - \frac{r}{4 + r}} = \frac{3}{2}$

$\Rightarrow (\frac{8}{8 + r}) * (\frac{4 + 8}{4}) = \frac{3}{2}$

$\Rightarrow 4 (4 + r) = 3 (8 + r)$

$\Rightarrow 16 + 4 r = 24 + 3 r$

r = 24 – 16

r = 8 $Ω$

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