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...
Read More →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...
Read More →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)....
Read More →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...
Read More →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, $x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$ $\Rightarrow 2(k+3)+4(-3-3)+6(3-k)=0$ $\Rightarrow 2 k+6-24+18-6 k=0$ $\Rightarrow-4 k=0$ $...
Read More →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...
Read More →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 $...
Read More →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 ...
Read More →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...
Read More →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...
Read More →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...
Read More →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 ...
Read More →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...
Read More →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...
Read More →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 ...
Read More →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...
Read More →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....
Read More →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, $n_{2}=\frac{25}{180 \mathrm{~g} \mathrm{~mol}^{-1}}$ = ...
Read More →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...
Read More →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, $P A=P B$ $\Rightarrow P A^{2}=P B^{2}$ $\Rightarrow(x-7)^{2}+(y-1)^{2}=(x-3)^{2}+(y-5)^{2}$ $\Rightarrow x^{2}+y^{2}-14 x-2 y+50=x^{2}+y^{2}-6 x-10 y+34$ $\Rightarrow 8 x-8 y=16$ $\Rightarrow x-y=2$...
Read More →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...
Read More →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{...
Read More →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...
Read More →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 $x=2$....
Read More →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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