Find the value of a, so that the point (3, a) lies on the line
Question: Find the value ofa, so that the point (3,a) lies on the line 2x 3y= 5. Solution: The point $(3, a)$ lies on the line $2 x-3 y=5$. If point $(3, a)$ lies on the line $2 x-3 y=5$, then $2 x-3 y=5$ $\Rightarrow(2 \times 3)-(3 \times a)=5$ $\Rightarrow 6-3 a=5$ $\Rightarrow 3 a=1$ $\Rightarrow a=\frac{1}{3}$ Hence, the value of $a$ is $\frac{1}{3}$....
Read More →Zener breakdown occurs in a p-n junction having p and n both :
Question: Zener breakdown occurs in a p-n junction having p and n both :lightly doped and have wide depletion layer.heavily doped andhave narrow depletion layer.heavily doped and have wide depletion layer.lightly doped and have narrow depletion layer.Correct Option: , 2 Solution: (2) The zener breakdown occurs in the heavily doped $\mathrm{p}-\mathbf{n}$ junction diode. Heavily doped $\mathrm{p}-\mathbf{n}$ junction diodes have narrow depletion region....
Read More →Calculate the depression in the freezing point of water when
Question: Calculate the depression in the freezing point of water when 10 g ofCH3CH2CHClCOOH is added to 250 g of water.Ka= 1.4 103,Kf= 1.86 K kg mol1. Solution: Molar mass of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCOOH}=15+14+13+35.5+12+16+16+1$ $=122.5 \mathrm{~g} \mathrm{~mol}^{-1}$ $\therefore$ No. of moles present in $10 \mathrm{~g}$ of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCOOH}=\frac{10 \mathrm{~g}}{122.5 \mathrm{~g} \mathrm{~mol}^{-1}}$ $=0.0816 \mathrm{~mol}$ It is given th...
Read More →Find the distance between the points
Question: Find the distance between the points $\left(\frac{-8}{5}, 2\right)$ and $\left(\frac{2}{5}, 2\right)$. Solution: The given points are $A\left(\frac{-8}{5}, 2\right)$ and $B\left(\frac{2}{5}, 2\right)$. Then, $\left(x_{1}=\frac{-8}{5}, y_{1}=2\right)$ and $\left(x_{2}=\frac{2}{5}, y_{2}=2\right)$ Therefore, $A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ $=\sqrt{\left\{\frac{2}{5}-\left(\frac{-8}{5}\right)\right\}^{2}+(2-2)^{2}}$ $=\sqrt{(2)^{2}+(0)^{2}}$ $=\sqrt{...
Read More →In connection with the circuit drawn below,
Question: In connection with the circuit drawn below, the value of current flowing through $2 \mathrm{k} \Omega$ resistor is______ $\times 10^{-4} \mathrm{~A}$ Solution: $(25)$ In zener diode there will be o change in current after $5 \mathrm{~V}$ zener diode breakdown $\Rightarrow i=\frac{5}{2 \times 10^{3}}$ $\Rightarrow \mathrm{i}=2.5 \times 10^{-3} \mathrm{~A}$ $\Rightarrow \mathrm{i}=25 \times 10^{-4} \mathrm{~A}$...
Read More →Find the point on x-axis which is equidistant from points A(−1, 0) and B(5, 0).
Question: Find the point onx-axis which is equidistant from pointsA(1, 0) andB(5, 0). Solution: LetP(x, 0) be the point onx-axis. Then $A P=B P \Rightarrow A P^{2}=B P^{2}$ $\Rightarrow(x+1)^{2}+(0-0)^{2}=(x-5)^{2}+(0-0)^{2}$ $\Rightarrow x^{2}+2 x+1=x^{2}-10 x+25$ $\Rightarrow 12 x=24 \Rightarrow x=2$ Hence,x= 2....
Read More →If an emitter current is changed by 4 mA,
Question: If an emitter current is changed by $4 \mathrm{~mA}$, the collector current changes by $3.5 \mathrm{~mA}$. The value of $\beta$7$0.875$$0.5$$3.5$Correct Option: 1 Solution: (1) Given : $\Delta \mathrm{I}_{\mathrm{E}}=4 \mathrm{~mA}$ $\Delta \mathrm{I}_{\mathrm{C}}=3.5 \mathrm{~mA}$ we know that, $\alpha=\frac{\Delta \mathrm{I}_{\mathrm{c}}}{\Delta \mathrm{I}_{\mathrm{E}}}$ $\Rightarrow \alpha=\frac{3.5}{4}=\frac{7}{8}$ Also, $\beta=\frac{\alpha}{1-\alpha}$, so $\beta=\frac{\frac{7}{8}}...
Read More →The depression in freezing point of water observed for the same amount of acetic acid,
Question: The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly. Solution: Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H+ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater...
Read More →If the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3 then find the value of k.
Question: If the pointC(k, 4) divides the join ofA(2, 6) andB(5, 1) in the ratio 2 : 3 then find the value ofk. Solution: Here, the pointC(k, 4) divides the join ofA(2, 6) andB(5, 1) in the ratio 2 : 3. So $k=\frac{2 \times 5+3 \times 2}{2+3}$ $=\frac{10+6}{5}$ $=\frac{16}{5}$ Hence, $k=\frac{16}{5}$....
Read More →Calculate the amount of benzoic acid
Question: Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol. Solution: 0.15 M solution of benzoic acid in methanol means, 1000 mL of solution contains 0.15 mol of benzoic acid Therefore, $250 \mathrm{~mL}$ of solution contains $=\frac{0.15 \times 250}{1000}$ mol of benzoic acid = 0.0375 mol of benzoic acid Molar mass of benzoic acid (C6H5COOH) = 7 12 + 6 1 + 2 16 = 122 g mol1 Hence, required benzoic acid = 0.0375 mol 122 g mol1 = 4.575 g...
Read More →Find the lengths of the medians AD and BE of
Question: Find the lengths of the medians $A D$ and $B E$ of $\Delta A B C$ whose vertices are $A(7,-3), B(5,3)$ and $C(3,-1)$. Solution: The given vertices areA(7, 3),B(5, 3) andC(3, 1).SinceDandEare the midpoints ofBCandACrespectively, therefore Coordinates of $D=\left(\frac{5+3}{2}, \frac{3-1}{2}\right)=(4,1)$ Coordinates of $E=\left(\frac{7+3}{2}, \frac{-3-1}{2}\right)=(5,-2)$ Now $A D=\sqrt{(7-4)^{2}+(-3-1)^{2}}=\sqrt{9+16}=5$ $B E=\sqrt{(5-5)^{2}+(3+2)^{2}}=\sqrt{0+25}=5$ Hence,AD=BE= 5 un...
Read More →Solve the following
Question: Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 103m aqueous solution required for the above dose. Solution: The molar mass of nalorphene $\left(\mathrm{C}_{19} \mathrm{H}_{21} \mathrm{NO}_{3}\right)$ is given as: $19 \times 12+21 \times 1+1 \times 14+3 \times 16=311 \mathrm{~g} \mathrm{~mol}^{-1}$ In 1.5 103m aqueous solution of nalorphene, $1 \mathrm{~kg}(1000...
Read More →Solve the following
Question: Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 103m aqueous solution required for the above dose. Solution: The molar mass of nalorphene $\left(\mathrm{C}_{19} \mathrm{H}_{21} \mathrm{NO}_{3}\right)$ is given as: $19 \times 12+21 \times 1+1 \times 14+3 \times 16=311 \mathrm{~g} \mathrm{~mol}^{-1}$ In 1.5 103m aqueous solution of nalorphene, $1 \mathrm{~kg}(1000...
Read More →The typical output characteristics curve for a transistor working in the common-emitter configuration is shown in the figure.
Question: The typical output characteristics curve for a transistor working in the common-emitter configuration is shown in the figure. The estimated current gain from the figure is Solution: (200) $\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{2 \times 10^{-3}}{10 \times 10^{-6}}$ $\beta=\frac{1}{5} \times 10^{3}$ $\beta=2 \times 10^{2}$ $\beta=200$...
Read More →Prove that the diagonals of a rectangle ABCD with vertices A(2, −1), B(5, −1), C(5, 6) and D(2, 6) are
Question: Prove that the diagonals of a rectangleABCDwith verticesA(2, 1),B(5, 1),C(5, 6) andD(2, 6) are equal and bisect each other. Solution: The vertices of the rectangleABCDareA(2, 1),B(5, 1),C(5, 6) andD(2, 6). Now Coordinates of midpoint of $A C=\left(\frac{2+5}{2}, \frac{-1+6}{2}\right)=\left(\frac{7}{2}, \frac{5}{2}\right)$ Coordinates of midpoint of $B D=\left(\frac{5+2}{2}, \frac{-1+6}{2}\right)=\left(\frac{7}{2}, \frac{5}{2}\right)$ Since, the midpoints ofACandBDcoincide, therefore th...
Read More →Calculate the mass percentage of aspirin
Question: Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4is dissolved in 450 g of CH3CN. Solution: 6.5 g of C9H8O4is dissolved in 450 g of CH3CN. Then, total mass of the solution = (6.5 + 450) g = 456.5 g Therefore, mass percentage of $\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}=\frac{6.5}{456.5} \times 100 \%$ = 1.424%...
Read More →Find the ratio in which the point P(x, 2) divides the join of A(12, 5) and B(4, −3).
Question: Find the ratio in which the pointP(x, 2) divides the join ofA(12, 5) andB(4, 3). Solution: Letkbe the ratio in which the pointP(x, 2) divides the line joining the pointsA(x1= 12,y1= 5) andB(x2= 4,y2= 3). Then $x=\frac{k \times 4+12}{k+1} \quad$ and $\quad 2=\frac{k \times(-3)+5}{k+1}$ Now $2=\frac{k \times(-3)+5}{k+1} \Rightarrow 2 k+2=-3 k+5 \Rightarrow k=\frac{3}{5}$ Hence, the required ratio is 3 : 5....
Read More →A square water tank has its side equal to 40 m.
Question: A square water tank has its side equal to 40 m. There are four semi-circular grassy plots all round it. Find the cost of turfing the plot at Rs. 1.25 per square metre (Take = 3.14). Solution: It is given that the side of square $a=40 \mathrm{~m}$ - Since four semicircular grassy plots rounds a square water tank. Then, diameter of semicircular plot is $2 r=a$. So, the radius of semicircle $r=\frac{a}{2}$ $=\frac{40}{2}$ $=20 \mathrm{~m}$ Area of semicircular plot $=\frac{1}{2} \pi r^{2}...
Read More →If the solubility product of CuS is
Question: If the solubility product of CuS is 6 1016, calculate the maximum molarity of CuS in aqueous solution. Solution: Solubility product of CuS,Ksp= 6 1016 Letsbe the solubility of CuS in mol L1. Now, =ss =s2 Then, we have, $K_{\mathrm{sp}}=s^{2}=6 \times 10^{-16}$ $\Rightarrow s=\sqrt{6 \times 10^{-16}}$ = 2.45 108mol L1 Hence, the maximum molarity of CuS in an aqueous solution is 2.45 108mol L1....
Read More →The correct relation between
Question: The correct relation between $\alpha$ (ratio of collector current to emitter current) and $\beta$ (ratio of collector current to base current) of a transistor is :$\beta=\frac{\alpha}{1+\alpha}$$\alpha=\frac{\beta}{1-\alpha}$$\beta=\frac{1}{1-\alpha}$$\alpha=\frac{\beta}{1+\beta}$Correct Option: , 4 Solution: (4) $\alpha=\frac{I_{C}}{I_{E}}, \beta=\frac{I_{C}}{I_{B}}$ $I_{E}=I_{B}+I_{C}$ $\alpha=\frac{I_{C}}{I_{B}+I_{C}}=\frac{1}{\frac{I_{B}}{I_{C}}+1}$ $\alpha=\frac{1}{\frac{1}{\beta}...
Read More →If the density of some lake water is
Question: If the density of some lake water is 1.25 g mL1and contains 92 g of Na+ions per kg of water, calculate the molality of Na+ions in the lake. Solution: Number of moles present in $92 \mathrm{~g}$ of $\mathrm{Na}^{+}$ions $=\frac{92 \mathrm{~g}}{23 \mathrm{~g} \mathrm{~mol}^{-1}}$ = 4 mol Therefore, molality of $\mathrm{Na}^{+}$ions in the lake $=\frac{4 \mathrm{~mol}}{1 \mathrm{~kg}}$ = 4 m...
Read More →If the point P(k − 1, 2) is equidistant from the points A(3, k) and B(k, 5),
Question: If the pointP(k 1, 2) is equidistant from the pointsA(3,k) andB(k, 5), find the values ofk. Solution: The given points areP(k 1, 2),A(3,k) andB(k, 5). $\because A P=B P$ $\therefore A P^{2}=B P^{2}$ $\Rightarrow(k-1-3)^{2}+(2-k)^{2}=(k-1-k)^{2}+(2-5)^{2}$ $\Rightarrow(k-4)^{2}+(2-k)^{2}=(-1)^{2}+(-3)^{2}$ $\Rightarrow k^{2}-8 y+16+4+k^{2}-4 k=1+9$ $\Rightarrow k^{2}-6 y+5=0$ $\Rightarrow(k-1)(k-5)=0$ $\Rightarrow k=1$ or $k=5$ Hence,k= 1 ork= 5....
Read More →Four equal circles, each of radius a, touch each other.
Question: Four equal circles, each of radius a, touch each other. Show that the area between them is $\frac{6}{7} a^{2}($ Take $\pi=22 / 7)$. Solution: It is given that four equal circles of radiusatouches each other. So, Area of circle $=\pi a^{2}$ Since circles touches each other, the lines joining their centre make a square ABCD. The side of square is 2a. Area of quadrant inside square $=\frac{1}{4} \pi a^{2}$ Area of shaded region $=$ Area of square $-4 \times$ Area of quadrant $=(2 a)^{2}-4...
Read More →Amongst the following compounds, identify which are insoluble,
Question: Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i)phenol(ii)toluene(iii)formic acid (iv)ethylene glycol(v)chloroform(vi)pentanol. Solution: (i)Phenol (C6H5OH) has the polar group OH and non-polar group C6H5. Thus, phenol is partially soluble in water. (ii)Toluene (C6H5CH3) has no polar groups. Thus, toluene is insoluble in water. (iii)Formic acid (HCOOH) has the polar group OH and can form H-bond with water. Thus, formic ac...
Read More →An npn transistor operates as a common emitter amplifier
Question: An npn transistor operates as a common emitter amplifier with a power gain of $10^{6}$. The input circuit resistance is $100 \Omega$ and the output load resistance is $10 \mathrm{~K} \Omega$. The common emitter current gain 'lbeta' will be (Round off to the Nearest Integer) Solution: $(100)$ $10^{6}=\beta^{2} \times \frac{\mathrm{R}_{0}}{\mathrm{R}_{\mathrm{i}}}$ $10^{6}=\beta^{2} \times \frac{10^{4}}{10^{2}}$ $\beta^{2}=10^{4} \Rightarrow \beta=100$...
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