eSaral - JEE | NEET | Class 8-10 Exam Prep App

In the following figure, OACB is a quadrant of a circle

Question: In the following figure,OACBis a quadrant of a circle with centreOand radius 3.5 cm. IfOD= 2 cm, find the area of the (i) quadrantOACB (ii) shaded region. Solution: It is given that OACB is a quadrant of circle with centre at O and radius 3.5 cm. (i) We know that the area of quadrant of circle of radiusris, $A=\frac{1}{4} \pi r^{2}$ Substituting the value of radius, $A=\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5$ $=9.625 \mathrm{~cm}^{2}$ Hence, the area of $\mathrm{OACB}$ is...

The air is a mixture of a number of gases.

Question: The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298Kif the Henrys law constants for oxygen and nitrogen are 3.30 107mm and 6.51 107mm respectively, calculate the composition of these gases in water. Solution: Percentage of oxygen (O2) in air = 20 % Percentage of nitrogen (N2) in air = 79% Also, it is given that water i...

The distance of the point P(−6, 8) from the origin is

Question: The distance of the pointP(6, 8) from the origin is (a) 8 (b) $2 \sqrt{7}$ (c) 6 (d) 10 Solution: The distance of a point $(x, y)$ from the origin $O(0,0)$ is $\sqrt{x^{2}+y^{2}}$. LetP(x= 6,y= 8) be the given point. Then $O P=\sqrt{x^{2}+y^{2}}$ $=\sqrt{(-6)^{2}+8^{2}}$ $=\sqrt{36+64}$ $=\sqrt{100}=10$ Hence, the correct answer is option (d)....

Benzene and toluene form ideal solution over the entire range of composition.

Question: Benzene and toluene form ideal solution over the entire range of composition.The vapour pressure of pure benzene and toluene at 300 K are 50.71mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene. Solution: Molar mass of benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=6 \times 12+6 \times 1$ $=78 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of toluene $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C...

If the points A(2, 3) B(4, k) and C(6, −3) are collinear, find the value of k.

Question: If the pointsA(2, 3)B(4,k) andC(6, 3) are collinear, find the value ofk. Solution: The given points are $A(2,3), B(4, k)$ and $C(6,-3)$ Here, $\left(x_{1}=2, y_{1}=3\right),\left(x_{2}=4, y_{2}=k\right)$ and $\left(x_{3}=6, y_{3}=-3\right)$ It is given that the points $A, B$ and $C$ are collinear. Then, $x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$ $\Rightarrow 2(k+3)+4(-3-3)+6(3-k)=0$ $\Rightarrow 2 k+6-24+18-6 k=0$ $\Rightarrow-4 k=0$ $...

LED is constructed from Ga-As-P semiconducting material. The energy

Question: LED is constructed from Ga-As-P semiconducting material. The energy gap of this LED is $1.9 \mathrm{eV}$. Calculate the wavelength of light emitted and its colour. $\left[\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\right.$ and $\left.\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right]$$654 \mathrm{~nm}$ and red colour$1046 \mathrm{~nm}$ and blue colour$1046 \mathrm{~nm}$ and red colour$654 \mathrm{~nm}$ and orange colourCorrect Option: 1 Solution: (1) We know that $\mathrm{E}=\frac{\ma...

In the following figure, AB and CD are two diameters of a

Question: In the following figure,ABandCDare two diameters of a circle perpendicular to each other andODis the diameter of the smaller circle. IfOA= 7 cm, find the area of the shaded region. Solution: It is given that AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of small circle. It is given that, $\mathrm{OA}=7 \mathrm{~cm}$' So, radiusrof small circle is $r=\frac{7}{2} \mathrm{~cm}$ $=3.5 \mathrm{~cm}$ We know that the area $A$ of circle of radius $...

In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7, 8)?

Question: In what ratio does the pointC(4, 5) divide the join ofA(2, 3) andB(7, 8)? Solution: Let the required ratio be $k: 1$. Then, by section formula, the coordinates ofCare $C\left(\frac{7 k+2}{k+1}, \frac{8 k+3}{k+1}\right)$ Therefore, $\frac{7 k+2}{k+1}=4$ and $\frac{8 k+3}{k+1}=5 \quad[\because C(4,5)$ is given $]$ $\Rightarrow 7 k+2=4 k+4$ and $8 k+3=5 k+5 \Rightarrow 3 k=2$ and $3 k=2$ $\Rightarrow k=\frac{2}{3}$ in each case So, the required ratio is $\frac{2}{3}: 1$, which is same as ...

For extrinsic semiconductors: when doping level is increased;

Question: For extrinsic semiconductors: when doping level is increased;Fermi-level of $p$ and $n$-type semiconductors will not be affected.Fermi-level of p-type semiconductors will go downward and Fermi-level of n-type semiconductor will go upward.Fermi-level of both p-type and n-type semiconductors will go upward for $TT_{F} K$ and downward for $\mathrm{T}\mathrm{T}_{\mathrm{F}} \mathrm{K}$, where $\mathrm{T}_{\mathrm{F}}$ is $\mathrm{Fermi}$ temperature.Fermi-level of p-type semiconductor will...

Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively.

Question: Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plotptotalpchloroform andpacetoneas a function ofxacetone. The experimental data observed for different compositions of mixture is. Solution: From the question, we have the following data It can be observed from the graph that the plot for theptotalof the solution curves downwards. Therefore, the solution show...

Find the area of the following figure,

Question: Find the area of the following figure, in square cm, correct to one place of decimal. (Take = 22/7). Solution: Let the area of square ABCD beA. It s given that, $\mathrm{AB}=10 \mathrm{~cm}$ So, $A=10 \times 10 \mathrm{~cm}^{2}$ It is given that a semicircle is attached to one side of the square. The diameter of semicircle $=10 \mathrm{~cm}$ So, radius $r$ of semicircle $=5 \mathrm{~cm}$ We know that the area of semicircle of radius r is $A^{\prime}=\frac{1}{2} \times \frac{22}{7} \tim...

Find the centroid of ∆ABC whose vertices are

Question: Find the centroid of ∆ABCwhose vertices areA(2, 2),B(4, 4) andC(5, 8). Solution: The given points areA(2, 2),B(4, 4) andC(5, 8). Here, $\left(x_{1}=2, y_{1}=2\right), \quad\left(x_{2}=-4, y_{2}=-4\right)$ and $\left(x_{3}=5, y_{3}=-8\right)$ Let $G(x, y)$ be the centroid of $\Delta A B C$. Then, $x=\frac{1}{3}\left(x_{1}+x_{2}+x_{3}\right)$ $=\frac{1}{3}(2-4+5)$ $=1$ $y=\frac{1}{3}\left(y_{1}+y_{2}+y_{3}\right)$ $=\frac{1}{3}(2-4-8)$ $=\frac{-10}{3}$ Hence, the centroid of $\Delta A B ...

100 g of liquid A

Question: 100 g of liquid A (molar mass 140 g mol1) was dissolved in 1000 g of liquid B (molar mass 180 g mol1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr. Solution: Number of moles of liquid $A, n_{\mathrm{A}}=\frac{100}{140} \mathrm{~mol}$ = 0.714 mol Number of moles of liquid $\mathrm{B}, n_{\mathrm{B}}=\frac{1000}{180} \mathrm{~m...

Choose the correct answer form the options given below:

Question: Choose the correct answer form the options given below:(a) $-$ (ii), (b) $-$ (i), (c) $-$ (iv), (d) $-$ (iii)$(a)-(i i),(b)-(i v),(c)-(i),(d)-(i i i)$$(\mathrm{a})-(\mathrm{ii}),(\mathrm{b})-(\mathrm{i}),(\mathrm{c})-(\mathrm{iii}),(\mathrm{d})-(\mathrm{i} v)$(a) $-$ (iii), (b) $-$ (iv), (c) $-$ (i), (d) $-$ (ii)Correct Option: , 2 Solution: (2) (a)Rectifier:- used to convert a.c voltage into d.c. Voltage. (b) Stabilizer:- used for constant output voltage even when the input voltage or...

If the centroid of ∆ABC, which has vertices A(a, b), B(b, c) and C(c, a),

Question: If the centroid of ∆ABC, which has verticesA(a,b),B(b,c) andC(c,a), is the origin, find the value of (a+b+c). Solution: The given points areA(a,b),B(b,c) andC(c,a).Here, $\left(x_{1}=a, y_{1}=b\right), \quad\left(x_{2}=b, y_{2}=c\right)$ and $\left(x_{3}=c, y_{3}=a\right)$ Let the centroid be(x, y).Then, $x=\frac{1}{3}\left(x_{1}+x_{2}+x_{3}\right)$ $=\frac{1}{3}(a+b+c)$ $=\frac{a+b+c}{3}$ $y=\frac{1}{3}\left(y_{1}+y_{2}+y_{3}\right)$ $=\frac{1}{3}(b+c+a)$ $=\frac{a+b+c}{3}$ But it is ...

The inside perimeter of a running track (shown in the following figure) is 400 m.

Question: The inside perimeter of a running track (shown in the following figure) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If the track is everywhere 14 m wide. find the area of the track. Also find the length of the outer running track. Solution: It is given that, length of each straight portion $=90 \mathrm{~m}$ and width of track $=14 \mathrm{~m}$ We know that the circumference C of semicircle of radius beris $C=\pi r$ The inside perimeter of...

Henry’s law constant for the molality of methane in benzene at 298 K is

Question: Henrys law constant for the molality of methane in benzene at 298 K is 4.27 105mm Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg. Solution: Here, p= 760 mm Hg kH= 4.27 105mm Hg According to Henrys law, p= kHx $\Rightarrow x=\frac{p}{k_{\mathrm{H}}}$ $=\frac{760 \mathrm{~mm} \mathrm{Hg}}{4.27 \times 10^{5} \mathrm{~mm} \mathrm{Hg}}$ = 177.99 105 = 178 105(approximately) Hence, the mole fraction of methane in benzene is 178 105....

Vapour pressure of water at 293 Kis 17.535 mm Hg.

Question: Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in 450 g of water. Solution: Vapour pressure of water, $p_{1}^{0}=17.535 \mathrm{~mm}$ of $\mathrm{Hg}$ Mass of glucose,w2= 25 g Mass of water,w1= 450 g We know that, Molar mass of glucose (C6H12O6),M2= 6 12 + 12 1 + 6 16 = 180 g mol1 Molar mass of water,M1= 18 g mol1 Then, number of moles of glucose, $n_{2}=\frac{25}{180 \mathrm{~g} \mathrm{~mol}^{-1}}$ = ...

A 5V battery is connected across the points X and Y.

Question: A $5 \mathrm{~V}$ battery is connected across the points $\mathrm{X}$ and $\mathrm{Y}$. Assume $\mathrm{D}_{1}$ and $\mathrm{D}_{2}$ to be normal silicon diodes. Find the current supplied by the battery if the tve terminal of the battery is connected to point $X$. $\sim 0.86 \mathrm{~A}$$\sim 0.5 \mathrm{~A}$$\sim 0.43 \mathrm{~A}$$\sim 1.5 \mathrm{~A}$Correct Option: , 3 Solution: (3) Since silicon diode is used so $0.7$ Volt is drop across it, only $D_{1}$ will conduct so current thr...

If P(x, y) is equidistant from the points

Question: IfP(x,y) is equidistant from the pointsA(7, 1) andB(3, 5), find the relation betweenxandy. Solution: Let the point $P(x, y)$ be equidistant from the points $A(7,1)$ and $B(3,5)$. Then, $P A=P B$ $\Rightarrow P A^{2}=P B^{2}$ $\Rightarrow(x-7)^{2}+(y-1)^{2}=(x-3)^{2}+(y-5)^{2}$ $\Rightarrow x^{2}+y^{2}-14 x-2 y+50=x^{2}+y^{2}-6 x-10 y+34$ $\Rightarrow 8 x-8 y=16$ $\Rightarrow x-y=2$...

Prove that the area of a circular path

Question: Prove that the area of a circular path of uniform widthhsurrounding a circular region of radiusris h(2r + h). Solution: We know that the area of a circle of radius $r$ is $A=\pi r^{2}$ It is given that a circular path of widthhsurrounds the circle of radiusr. So, radius of the outer circle $=r+h$ Using the value of radius in above formula, Area of the outer circle $=\pi(r+h)^{2}$ Hence, Area of the circular path = Area of outer circle-Area of inner circle $=\pi(r+h)^{2}-\pi r^{2}$ $=\p...

Solve the following

Question: 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0C. Calculate the vant Hoff factor and dissociation constant of fluoroacetic acid. Solution: It is given that: $w_{1}=500 \mathrm{~g}$ $w_{2}=19.5 \mathrm{~g}$ $K_{f}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ $\Delta T_{f}=1 \mathrm{~K}$ We know that: $M_{2}=\frac{K_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}}$ $=\frac{1.86 \mathrm{~K} \mathrm{~kg} \mathrm{...

Given below are two statements :

Question: Given below are two statements : Statement I : PN junction diodes can be used to function as transistor, simply by connecting two diodes, back to back, which acts as the base terminal. Statement II : In the study of transistor, the amplification factor $\beta$ indicates ratio of the collector current to the base current. In the light of the above statements, choose the correct answer from the options given below.Statement I is false but Statement II is true.Both Statement I and Stateme...

If the points A(4, 3) and B(x, 5) lie on the circle with centre O(2, 3),

Question: If the pointsA(4, 3) andB(x, 5) lie on the circle with centreO(2, 3), find the value ofx. Solution: The given points $A(4,3)$ and $B(x, 5)$ lie on the circle with centre $O(2,3)$. Then, $O A=O B$ $\Rightarrow \sqrt{(x-2)^{2}+(5-3)^{2}}=\sqrt{(4-2)^{2}+(3-3)^{2}}$ $\Rightarrow(x-2)^{2}+2^{2}=2^{2}+0^{2}$ $\Rightarrow(x-2)^{2}=\left(2^{2}-2^{2}\right)$ $\Rightarrow(x-2)^{2}=0$ $\Rightarrow x-2=0$ $\Rightarrow x=2$ Hence, the value of $x=2$....

A rectangular park is 100 m by 50 m.

Question: A rectangular park is 100 m by 50 m. It is surrounding by semi-circular flower beds all round. Find the cost of levelling the semi-circular flower beds at 60 paise per square metre (use = 3.14). Solution: Since four semicircular flower beds rounds the rectangular park. Then, diameters of semicircular plots are $2 r_{1}=l$ and $2 r_{2}=w$ So, the radius of semicircle at larger side of rectangle $r_{1}=\frac{l}{2}$ $=\frac{100}{2}$ $=50 \mathrm{~m}$ Area of semicircluar plot at larger si...

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