Exemplar Solution

This is a multiple choice answer as classified in NCERT Exemplar

option  iii. p₁  < p₂ < p₃ < p₄

At a particular temperature, PV is constant

Therefore,   V ∝ $\frac{1}{\mathit{p}}$

So, as v₁ >  v₂  >  v₃  >  v₄   the order of pressure:  p₁  < p₂ < p₃ < p₄ .

This is a multiple choice answer as classified in NCERT Exemplar

option (ii) surface tension 

Surface tension minimises the surface area of the liquid and at  minimum surface area, the liquid is in its the lowest energy state and hence most stable. Spherical shape of rain droplets has minimum surface area due to surface tension. 

This is a multiple choice answer as classified in NCERT Exemplar

Option (iii) pressure decreases 

At high altitude the pressure is low and therefore, water boils at lower temperature because the vapour pressure becomes equal to atmospheric temperature at low temperature due to which the food takes more time to be cooked without using a pressure cooker.

This is a  Fill in the Blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{-\pi }{\overset{\pi }{\int }}si{n}^{3}x.co{s}^{2}x}dx\\ Letf\left(x\right)=si{n}^{3}x.co{s}^{2}x\\ f\left(-x\right)=si{n}^3\left(-x\right).co{s}^2\left(-x\right)=-si{n}^{3}x.co{s}^{2}x=-f\left(x\right)\\ \therefore {\displaystyle \underset{-\pi }{\overset{\pi }{\int }}si{n}^{3}x.co{s}^{2}x}dxisanoddfunction.\\ \therefore I=0\end{array}$

This is a  Fill in the Blanks type Questions as classified in NCERT Exemplar

This is a  Fill in the Blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{a}{\int }}\frac{1}{1+4x^2}}dx=\frac{\pi }{8}\\ \Rightarrow \frac{1}{4}{\displaystyle \underset{0}{\overset{a}{\int }}\frac{1}{\left(\frac{1}{4}+x^2\right)}}dx=\frac{\pi }{8}\Rightarrow {\displaystyle \underset{0}{\overset{a}{\int }}\frac{1}{\left[{\left(\frac{1}{2}\right)}^2+x^2\right]}}dx=\frac{\pi }{2}\\ \Rightarrow \frac{1}{1/2}{\left[ta{n}^{-1}\frac{x}{1/2}\right]}_0^a=\frac{\pi }{2}\Rightarrow 2\left[ta{n}^{-1}2a-ta{n}^{-1}0\right]=\frac{\pi }{2}\\ \Rightarrow ta{n}^{-1}2a=\frac{\pi }{4}\Rightarrow 2a=tan\frac{\pi }{4}\Rightarrow 2a=1\Rightarrow a=\frac{1}{2}\\ Hence,thevalueofa=\frac{1}{2}.\end{array}$

This is a  Fill in the Blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{x+3}{{\left(x+4\right)}^2}.e^x dx}\\ ={\displaystyle \int \frac{x+4-1}{{\left(x+4\right)}^2}.e^x dx}\\ ={\displaystyle \int \left[\frac{x+4}{{\left(x+4\right)}^2}-\frac{1}{{\left(x+4\right)}^2}\right].e^x dx}={\displaystyle \int \left[\frac{1}{x+4}-\frac{1}{{\left(x+4\right)}^2}\right].e^x dx}\\ Put\frac{1}{x+4}=t\Rightarrow -\frac{1}{{\left(x+4\right)}^2}dx=dt\\ Letf\left(x\right)=\frac{1}{x+4}\therefore f^{\text{'}}\left(x\right)=-\frac{1}{{\left(x+4\right)}^2}\\ Using{\displaystyle \int e^x\left[f\left(x\right)+f^{\text{'}}\left(x\right)\right]dx=}{e}^{x}f\left(x\right)+C\\ \therefore I=e^x.\frac{1}{x+4}+C\\ Hence,I=\frac{e^x}{x+4}+C.\end{array}$

This is a  Fill in the Blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cosx.e^{sinx}}dx\\ Putsinx=t\Rightarrow cosxdx=dt\\ Whenx=0thent=sin0=0Whenx=\frac{\pi }{2}thent=sin\frac{\pi }{2}=1\\ \therefore I={\displaystyle \underset{0}{\overset{1}{\int }}{e}^{t}dt}={\left[e^t\right]}_0^1=\left(e^1-e^0\right)=e-1\\ Hence,I=e-1\end{array}$

This is a Objective answer type Questions as classified in NCERT Exemplar