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

Two objects of equal masses placed at certain from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other, then the new force will be:

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alok kumar singh

Contributor-Level 10

F = $\frac{G m_{2}}{d^{2}}$ - (i)

Now, $F' = \frac{G m_{1} m_{2}}{d^{2}} = \frac{G 2 m}{3} * \frac{4 m}{3 d^{2}}$

$k' = \frac{8}{9} G \frac{m^{2}}{d^{2}}$

$∴ F' = \frac{8}{G} F$

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

The reagent, from the following, which converts benzoic acid to benzaldehyde in one step is

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Six

Water falls from a 40m height dam at the rate of 9 * 10⁴ kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydro electric energy number of 100 W lamps, that can be lit, is:

(Take g = 10 ms^-2)

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alok kumar singh

Contributor-Level 10

Given : water flow rate = $(\overset{o}{m})$ = 9 * 10⁴kg/hr

= $\frac{9 * 1 0^{4} k g / s e c}{60 * 60}$

Potential energy available on water per unit time

$\Rightarrow \overset{o}{m} g h = \frac{9 * 1 0^{4}}{60 * 60} * 10 * 40$

$\overset{o}{m} g h = 1 0^{4} w a t t$

50% of it converts into electrical energy = 50% of $\overset{o}{m} g h = \frac{1 0^{4}}{2} w a t t$

Let $η$ be the number of bulbs n = 100 = $\frac{1 0^{4}}{2}$

n = 50

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Halogenation of which one of the following will yield m-substituted product with respect to methyl group as a major product?

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Vishal Baghel

Contributor-Level 10

OH group is the more activating group and ortho-para director, hence ortho position attach will provide meta substituted product with respect to – CH₃ in option (C).

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A $\sqrt[]{34} m$ long ladder weighing 10kg leans on a frictionless wall. Its feet rest on the floor 3m away from the wall as shown in the figure. If F_f and F_w are the reaction forces of the floor and the wall, then ratio of F_w / F_f will be:

(Use g = 10 m/s²)

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alok kumar singh

Contributor-Level 10

mg = 10 * 10

= 100 of

NA = $F_{ω}$ reaction force offered by the wall

$\sqrt[]{N_{3}^{2} + f_{B}^{2}} = F_{f}$ reaction force offered by the floor.

By NLM 1

Translational fB = NA

NB = mg = 100N

Ac² + Bc² = AB²

Ac² = 34 – 9

Ac = 5m

Rotational equilibrium $τ_{B} = 0$

mg $\frac{l}{2} c o s θ = N_{A} l s i n θ$

$100 \frac{c o s θ}{2} = N_{A} s i n θ$

NA = 50 cot

Now cot = $\frac{3}{5}$

N_A= $50 * \frac{3}{5} = 30 N$

Therefore, NA = $F_{ω} = 30 N$ - (i)

$F_{f} = \sqrt[]{N_{B}^{2} + f_{B}^{2}} = \sqrt[]{10 0^{2} + 3 0^{2}} = 10 \sqrt[]{109} N$ - (2)

$\frac{f_{ω}}{F_{f}} = \frac{30}{10 \sqrt[]{109}} = \frac{3}{109}$

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

The major product in the following reaction

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Oxidation of toluene to benzaldehyde can be easily carried out with which of the following

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

The correct order of nucleophilicity is

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Vishal Baghel

Contributor-Level 10

Nucleophilicity is directly proportional to the electron density of atom and inversely proportional to its electronegativity.

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Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Six

A block of mass 2kg moving on a horizontal surface with speed of 4 ms^-¹ enters a rough surface ranging from x = 0.5m to x = 1.5m. The retarding force in this range of rough surface is related to distance by F = -kx where k = 12Nm^-¹. The speed of the block as it just crosses the rough surface will be:

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alok kumar singh

Contributor-Level 10

by, WET $\int F d x = Δ k . 6$

$\int - k x d x = \frac{1}{2} m [v_{t}^{2} - v_{i}^{2}]$

$\Rightarrow \frac{- k}{2} {[x_{f}^{2} - x_{i}^{2}]}_{0.5}^{1.5} = \frac{1}{2} * 2 [v_{f}^{2} - 16]$ $[? v_{i} = 4 m / s e c]$

$\frac{- 12}{2} [1 . 5^{2} - 0 . 5^{2}] = [v_{f}^{2} - 16]$

$v_{f}^{2} = 16 - 12 = 4$

vf = 2m/sec

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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

The measured BOD values for four different water sample (A-D) are as follows

A = 3 ppm; B = 18 ppm: C = 21 ppm; D = 4 ppm.

The water samples which can be called as highly polluted with organic wastes, are

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Vishal Baghel

Contributor-Level 10

Highly polluted water has higher value of BOD $\geq$ 17 ppm

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