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Physics NCERT Exemplar Solutions Class 12th Chapter Five

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Five Surface Chemistry

Consider the ions /molecule $O_{2}^{+}, O_{2}, O_{2}^{-}, O_{2}^{2 -}$

For increasing bond order the correct option is

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Payal Gupta

Contributor-Level 10

$B o n d o r d e r = \frac{e^{-} o f B o n d i n g M O - e^{-} o f A B M O}{2}$

Hence Bond order for $O_{2}^{+} = \frac{10 - 5}{2} = 2.5$

$O_{2} = \frac{10 - 6}{2} = 2.0$

$O_{2}^{-} = \frac{10 - 7}{2} = 1.5$

$O_{2}^{- -} = \frac{10 - 8}{2} = 1.0$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Five Laws of Motion

A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is __________m.

[Use g = 10m/s²]

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Payal Gupta

Contributor-Level 10

Let 'h' be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

$h = u t + \frac{1}{2} a t^{2}$

$= 0 * 1 + \frac{1}{2} * 10 * 1 * 1$

= 5m

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Physics NCERT Exemplar Solutions Class 12th Chapter Five Laws of Motion

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads^-1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the will then rotate with an angular velocity (in rads^-1).

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Payal Gupta

Contributor-Level 10

By conservation of Angular momentum

L_i = L_f

$M R^{2} ω = (m R^{2} + 2 m R^{2}) ω'$

$\frac{2 M}{M + 2 m} = ω'$

$\Rightarrow$ $ω' = \frac{2 M}{M + 2 m}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Five Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

For the network shown below, the value of V_B – VA is_________ V.

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alok kumar singh

Contributor-Level 10

I= $\frac{15}{3} = 5 A$

$V_{B} + I * 1 ? 15 = v_{A}$

$v_{B} ? v_{A} = 15 ? 5 * 1$

= 10v

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Five Ray Optics and Optical Instruments

A parallel beam of light is allowed to afall on a transparent spherical globe of diameter 30cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is_________ mm.

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alok kumar singh

Contributor-Level 10

When object is at infinity

u = $\infty, μ_{1} = 1, R = 15 c m$

v = ? $μ_{2} = 1.5$

$\frac{- μ_{1}}{u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{- 1}{\infty} + \frac{1.5}{v} =$ $\frac{1.5 - 1}{15}$

$\frac{1.5}{v} =$ $\frac{0.5}{15}$

v =

45cm

Now for Refraction through C2

u = + 15cm

v = ?

$μ_{1} = 1.5$

$μ_{2} = 1$

R = 15

$\frac{- μ_{1}}{u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{- 1.5}{15} + \frac{1}{v} = \frac{1 - 1.5}{- 15}$

$\frac{- 1}{10} + \frac{1}{v} = \frac{1}{30}$

$\frac{1}{v} = \frac{1}{30} + \frac{1}{10} = \frac{4}{10}$

$v = \frac{30}{4} = + 7.5 c m$

from centre = 15 + 7.5 = 22.5 cm = 225mm

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Physics NCERT Exemplar Solutions Class 12th Chapter Five Semiconductor Electronics

A transistor is used in an amplifier circuit in common emitter mode. If the best current changes by 100 $μ A,$ it brings a change of 10 mA is collector current. If the load resistance is 2 $k Ω$ and input resistance is 1 $k Ω,$ the value of power gain is x * 10⁴. The value of x is_____________.

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alok kumar singh

Contributor-Level 10

Power gain = ${(\frac{Δ i_{c}}{Δ i_{b}})}^{2} * \frac{R_{0}}{R_{i}}$

= ${(\frac{10 * 1 0^{- 3}}{100 * 1 0^{- 6}})}^{2} * \frac{2}{1}$

= 2 * 104 = x * 104

= 2

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Physics NCERT Exemplar Solutions Class 12th Chapter Five Physics Nuclei

$\sqrt[]{d_{1}} a n d \sqrt[]{d_{2}}$ are the impact parameters corresponding to scattering angles 60° and 90° respectively, when an a particle is approaching a gold nucleus. For d₁ = x d₂, the value of x will be________.

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alok kumar singh

Contributor-Level 10

$\sqrt[]{d}$ $c o t \frac{θ}{2}$

cot² 30° = x cot² 45°

x = 3

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Physics NCERT Exemplar Solutions Class 12th Chapter Five Thermodynamics

300 cal. Of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227°C, then the temperature of sink will be__________°C.

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alok kumar singh

Contributor-Level 10

T₁ = 227°C

= 500k

T₂ =?

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

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

...more

T₁ = 227°C

= 500k

T₂ =?

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

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

T₂ = 375k

T₂ = 102°C

less

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Physics NCERT Exemplar Solutions Class 12th Chapter Five Physics NCERT Exemplar Solutions Class 12th Chapter Three

The variation of applied potential and flowing through a given wire is shown in figure. The length of wire is 31.4cm. The diameter of wire is measured as 2.4cm. The resistivity of the given wire is measured as x * 10^-3 $Ω c m .$ The value of x is_________.

[Take p = 3.14]

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alok kumar singh

Contributor-Level 10

R = tanq = tan 45° = $\frac{ρ L}{A}$

$? 1 = ρ * \frac{31.4}{π {(1.2)}^{2}}$

$ρ = \frac{3.14 * 1.2 * 1.2}{31.4}$

$ρ = 12 * 12 * 1 0^{- 3}$

$ρ = 144 * 1 0^{- 3} = x * 1 0^{- 3} Ω - c m$

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Physics NCERT Exemplar Solutions Class 12th Chapter Five Physics Oscillations

A body is performing simple harmonic with an amplitude of 10cm. The velocity if the body was tripled by air Jet when it is at 5cm from its mean position. The new amplitude vibration is $\sqrt[]{x} c m .$ The value of x is__________.

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alok kumar singh

Contributor-Level 10

v = $ω \sqrt[]{A^{2} - x^{2}}$ $^{}^{}$

at x = 5, A = 10

$v' = 3 v = 3 ω \sqrt[]{A^{2} - 5^{2}} = ω \sqrt[]{A'^{2} - 5^{2}}$

$= 3 \sqrt[]{A^{2} - 5^{2}} = \sqrt[]{A'^{2} - 5^{2}}$

$\sqrt[]{1 0^{2} - 5^{2}} = \sqrt[]{A'^{2} - 25}$

$A'^{2} = 25 + 9 * 75$ $A'^{2} = 700$

$A' = \sqrt[]{700} c m$

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