Tags Class 12th

Class 12th

Get insights from 11.9k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th

Follow Ask Question

11.9k

Questions

Discussions

Active Users

Followers

New answer posted

11 months ago

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?

0 Follower 4 Views

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).

0 0

New answer posted

11 months ago

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)

0 Follower 18 Views

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)

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

0 0

New answer posted

11 months ago

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

The major product in the following reaction

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

0 0

New answer posted

11 months ago

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

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

0 0

New answer posted

11 months ago

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

The correct order of nucleophilicity is

0 Follower 1 View

Vishal Baghel

Contributor-Level 10

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

0 0

New answer posted

11 months ago

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:

0 Follower 2 Views

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

0 0

New answer posted

11 months ago

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

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

0 0

New answer posted

11 months ago

Class 12th Polymers

${(C_{7} H_{5} O_{2})}_{2} \overset{h v}{\to} [X] \to 2 {\overset{•}{C}}_{6} H_{5} + 2 C O_{2}$

Consider the above reaction and identify the intermediate 'X'

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

This is the formation of free radical mechanism, hence free radical will be formed

0 0

New answer posted

11 months ago

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

A ball is spun with angular acceleration a = 6t² – 2t where t is in second and a is in rads^-¹. At t = 0, the ball has angular velocity of 10 rads^-¹ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

a = 6t² – 2t, of t 20, w = 10 rad/sec

, q = 4 rad

$α = \frac{d ω}{d t}$

$\Rightarrow \int_{10}^{ω} d ω = \int_{t - 0}^{t} α d t$

$\Rightarrow ω - 10 = \int_{0}^{t} (6 t^{2} - 2 t) d t$

$ω = \frac{d θ}{d t}$

$d θ = ω d t$

$\int_{4}^{θ} d θ = \int_{0}^{t} (2 t^{3} - t^{2} + 10) d t$

= $\frac{t^{4}}{2} - \frac{t^{3}}{3} + 10 t + 4$

0 0

New answer posted

11 months ago

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

The most common oxidation state of Lanthanoid elements is +3. Which of the following is likely to deviate easily from +3 oxidation state?

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$C e^{3 +} \to [X e] 4 f^{1} 5 d^{0}$

$C e^{4 +} \to [X e] 4 f^{0} 5 d^{0}$ (Noble gas configuration)

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
703k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 12th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience