Class 11th

This is a Short Answers Type Questions as classified in NCERT Exemplar

As we move down the group in group 13 and 14 the participation of s-electrons in bond formation decreases the primary reason behind this is the inert pair effect.

In this the p-electrons take part in bond formation and more energy is required to unpack the valence electrons to make them participate in bonding. Due to this the lower oxidation state of elements becomes stable done the hire oxidation state. As for group 13, +1 oxidation state is more stable than +3 and for group 14, +2 oxidation state is more stable than +4.

This is a long answer type question as classified in NCERT Exemplar

Frictional force f is acting in the opposite direction of F . let the acceleration of centre of mass of disc be a then

F-f=Ma where M is the mass of the disc

$\alpha =\frac{a}{R}$

$\tau =I\alpha$

fR= (1/2 MR²) $\alpha$

so fR= (1/2MR²) (a/R)

Ma=2f

From the above equation

F = F/3

F< $\mu N$ = $\mu Mg$

f/3 < $\mu Mg$

F=3 $\mu Mg$

This is a long answer type question as classified in NCERT Exemplar

area of square = area of rectangular plate

C²=a $*$ b

(a) $\frac{I_{xR}}{I_{xS}}=\frac{b^2}{c^2}$ here bxRxS

(b) $\frac{I_{yR}}{I_{yS}}=\frac{a^2}{c^2}$ here a>c so I_yR>I_yS

(c) I_zR-I_{zS $\propto \left(a^2+b^2-2c^2\right)=a^2+b^2$} +2ab= (a-b)²

(I_zR-I_zS)>0

$\frac{I_{zR}}{I_{zS}}1$

This is a Short Answers Type Questions as classified in NCERT Exemplar

Both BCl₃ and AlCl₃ are electron deficient compounds that are central atom boron and aluminium have incomplete Octet. In each compound, a metal atom is surrounded by six electrons of three covalent bonds with 3 chlorine atoms.

Each chlorine atom has a complete Octet of eight electrons. The electron deficient compounds act as Lewis acid and readily accept two electrons to complete their octet.

This is a long answer type question as classified in NCERT Exemplar

(a) Situation as given below

(b) F' =F=F'' where F' and F'' are external forces through support.

So F_net=O

External torque = F $*3R$ (anticlockwise)

(c) Let w₁ and w₂ be the final angular velocities of smaller and bigger drum in both clockwise and anticlockwise . finally there will be no friction

hence Rw₁=2Rw₂

so $\frac{w_1}{w_2}=2$

This is a Short Answers Type Questions as classified in NCERT Exemplar

H₃BO₃

Boric acid forms a hexagon and rings through hydrogen bonding and has a layer-like structure.

Boric acid is present in water as [B (OH)₄]−.H₃BO₃ electron from the OH of water and forms the complex BOH for negative for sp³ and is present in sp³ hybridisation.

Reaction:

B (OH)₃+2H₂O→ [B (OH)₄]⁻ + H₃O⁺

This is a long answer type question as classified in NCERT Exemplar

(a) Before being bought in contact with the table the disc was in pure rotational motion. Hence V_cm=0

(b) when the disc is placed in contact with table due to friction centre of mass acquires some linear velocity.

(c) when the rotating disc is placed in contact with the table due to friction centre of mass acquires some linear velocity.

(d) friction is responsible for the effects ib b and c

(e) when rolling starts V_cm=wR

(f) time period for rolling to begin is t= $\frac{R{w}_o}{\mu g(1+\frac{m{R}^2}{I})}$

This is a long answer type question as classified in NCERT Exemplar

(a) yes the law of conservation of angular momentum can be applied because there is no net external torque on the system of the two discs.

External forces gravitation and normal reaction act through the axis of rotation, hence produce no torque.

(b) by conservation of angular momentum

L_f= L_i

Iw=I₁w₁+I₂w₂

So w= $\frac{I_1{w}_1+I_2{w}_2}{I_1+I_2}$

(c) K_f= $\frac{1}{2}(I_1+I_2)\frac{{(I_1{w}_1+I_2{w}_{2)}}^2}{I_1+I_2}$

K_f= ½ (I₁w₁²+I₂w₂²)

$?k=$ K_f-K_i =- $\frac{-I_1{I}_2}{2\left(I_1+I_2\right)}(w_1-w_2)$ ²<0

(d) hence there is loss of KE of the system. The loss of kinetic energy is mainly due to work against the friction.