In the following figure, O is the centre of a circular arc and AOB is a straight line.
Question: In the following figure,Ois the centre of a circular arc andAOBis a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take = 3.142) Solution: (i) Let us find the perimeter of the shaded region. $\therefore$ Perimeter $=\pi \times 10+12+16$ $\therefore$ Perimeter $=3.142 \times 10+28$ $\therefore$ Perimeter $=31.42+28$ $\therefore$ Perimeter $=59.42$ Therefore, perimeter of the shaded region is $59.4 \mathrm{~cm}$ Now we will find the are...
Read More →In the following figure, there are three semicircles,
Question: In the following figure, there are three semicircles, A, B and C having diameter 3 cm each, and another semicircle E having a circle D with diameter 4.5 cm are shown. Calculate:(i) the area of the shaded region(ii) the cost of painting the shaded region at the rate of 25 paise per cm2, to the nearest rupee. Solution: (i) Area of the shaded region can be calculated as shown below, Area of the shaded region = Area of the semi-circle with diameter of 9 cm area of 2 semi-circles with radiu...
Read More →The points A(−4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle, which is
Question: The pointsA(4, 0),B(4, 0) andC(0, 3) are the vertices of a triangle, which is(a) isosceles(b) equilateral(c) scalene(d) right-angled Solution: (a) isoscelesLetA(4, 0),B(4, 0) andC(0, 3) be the given points. Then, $A B=\sqrt{(4+4)^{2}+(0-0)^{2}}$ $=\sqrt{(8)^{2}+(0)^{2}}$ $=\sqrt{64+0}$ $=\sqrt{64}$ $=8$ units $B C=\sqrt{(0-4)^{2}+(3-0)^{2}}$ $=\sqrt{(-4)^{2}+(3)^{2}}$ $=\sqrt{16+9}$ $=\sqrt{25}$ $=5$ units $A C=\sqrt{(0+4)^{2}+(3-0)^{2}}$ $=\sqrt{(4)^{2}+(3)^{2}}$ $=\sqrt{16+9}$ $=\sqr...
Read More →Predict the products of electrolysis in each of the following:
Question: Predict the products of electrolysis in each of the following: (i)An aqueous solution of AgNO3with silver electrodes. (ii)An aqueous solution of AgNO3with platinum electrodes. (iii)A dilute solution of H2SO4with platinum electrodes. (iv)An aqueous solution of CuCl2with platinum electrodes. Solution: (i)At cathode: The following reduction reactions compete to take place at the cathode. $\mathrm{Ag}_{(a q)}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}_{(s)} ; E^{0}=0.80 \mathrm{~V}$ $\...
Read More →In the following figure, ABC is a right-angled triangle,
Question: In the following figure,ABCis a right-angled triangle, B= 90,AB= 28 cm andBC= 21 cm. WithACas diameter a semicircle is drawn and withBCas radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places. Solution: We have given two semi-circles and one circle. Area of the shaded region = area of semicircle with diameter AC + area of right angled triangle ABC area of sector First we will find the hypotenuse of right angled triangle ABC. $A C^{2}=A B^{2}...
Read More →In the following figure, shows the cross-section of railway tunnel.
Question: In the following figure, shows the cross-section of railway tunnel. The radiusOAof the circular part is 2 m. If AOB= 90, calculate:(i) the height of the tunnel(ii) the perimeter of the cross-section(iii) the area of the cross-section. Solution: We have a cross section of a railway tunnel.is a right angled isosceles triangle, right angled at O. let OM be perpendicular to AB. (i) We have to find the height of the tunnel. We have, $\mathrm{OA}=2 \mathrm{~m}$ Use Pythagoras theorem in $\tr...
Read More →Two vertices of ∆ABC are A(−1, 4) and B(5, 2) and its centroid is G(0, −3)
Question: Two vertices of ∆ABCareA(1, 4) andB(5, 2) and its centroid isG(0, 3). Then, the coordinates ofCare(a) (4, 3)(b) (4, 15)(c) (4, 15)(d) (15, 4) Solution: (c) (4, 15) Two vertices of $\triangle A B C$ are $A(-1,4)$ and $B(5,2)$. Let the third vertex beC(a, b).Then, the coordinates of its centroid are $G\left(\frac{-1+5+a}{3}, \frac{4+2+b}{3}\right)$ i.e. $G\left(\frac{4+a}{3}, \frac{6+b}{3}\right)$ But it is given that the centroid is $G(0,-3)$. Therefore, $\frac{4+a}{3}=0$ and $\frac{6+b...
Read More →Predict the products of electrolysis in each of the following:
Question: Predict the products of electrolysis in each of the following: (i)An aqueous solution of AgNO3with silver electrodes. (ii)An aqueous solution of AgNO3with platinum electrodes. (iii)A dilute solution of H2SO4with platinum electrodes. (iv)An aqueous solution of CuCl2with platinum electrodes. Solution: (i)At cathode: The following reduction reactions compete to take place at the cathode. $\mathrm{Ag}_{(a q)}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}_{(s)} ; E^{0}=0.80 \mathrm{~V}$ $\...
Read More →Using the standard electrode potentials given in Table 3.1,
Question: Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: (i)Fe3+(aq) and I(aq) (ii)Ag+(aq) and Cu(s) (iii)Fe3+(aq) and Br(aq) (iv)Ag(s) and Fe3+(aq) (v)Br2(aq) and Fe2+(aq). Solution: Sincefor the overall reaction is positive, the reaction between Fe3+(aq)and I(aq)is feasible. Sincefor the overall reaction is positive, the reaction between Ag+(aq)and Cu(s)is feasible. Sincefor the overall reaction is negative, the reaction b...
Read More →A path of 4 m width runs round a semi-circular grassy plot
Question: A path of $4 \mathrm{~m}$ width runs round a semi-circular grassy plot whose circumference is $163 \frac{3}{7} \mathrm{~m}$ Find: (i) the area of the path(ii) the cost of gravelling the path at the rate of Rs 1.50 per square metre (iii) the cost of turfing the plot at the rate of 45 paise per $\mathrm{m}^{2}$. Solution: We have given $\mathrm{AB}=4 \mathrm{~m}$ and circumference of semicircle with radius $\mathrm{OA}$ as $163 \frac{3}{7} \mathrm{~m}$. We are asked to find the area betw...
Read More →If A(−1, 0), B(5, −2) and C(8, 2) are the vertices of a ∆ABC, then its centroid is
Question: IfA(1, 0),B(5, 2) andC(8, 2) are the vertices of a ∆ABC,then its centroid is(a) (12, 0)(b) (6, 0)(c) (0, 6)(d) (4, 0) Solution: (d) (4, 0) The given points are $A(-1,0), B(5,-2)$ and $C(8,2)$. Here, $\left(x_{1}=-1, y_{1}=0\right),\left(x_{2}=5, y_{2}=-2\right)$ and $\left(x_{3}=8, y_{3}=2\right)$ Let $G(x, y)$ be the centroid of $\Delta A B C$. Then, $x=\frac{1}{3}\left(x_{1}+x_{2}+x_{3}\right)$ $=\frac{1}{3}(-1+5+8)$ $=4$ and $y=\frac{1}{3}\left(y_{1}+y_{2}+y_{3}\right)$ $=\frac{1}{3...
Read More →Three electrolytic cells A,B,C containing solutions of
Question: Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? Solution: According to the reaction: i.e., 108 g of Ag is deposited by 96487 C. Therefore, $1.45 \mathrm{~g}$ of $\mathrm{Ag}$ is deposited by $=\frac{96487 \times 1.45}{108} \mathrm{...
Read More →Solve the following
Question: A solution of Ni(NO3)2is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode? Solution: Given, Current = 5A Time = 20 60 = 1200 s Charge = current time = 5 1200 = 6000 C According to the reaction, Nickel deposited by 2 96487 C = 58.71 g Therefore, nickel deposited by $6000 \mathrm{C}=\frac{58.71 \times 6000}{2 \times 96487} \mathrm{~g}$ = 1.825 g Hence, 1.825 g of nickel will be deposited at the cathode....
Read More →If A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC and AD is median, then the coordinates of D are
Question: IfA(4, 2),B(6, 5) andC(1, 4) be the vertices of ∆ABCandADis median, then the coordinates ofD are (a) $\left(\frac{5}{2}, 3\right)$ (b) $\left(5, \frac{7}{2}\right)$ (C) $\left(\frac{7}{2}, \frac{9}{2}\right)$ (d) None of these Solution: (c) $\left(\frac{7}{2}, \frac{9}{2}\right)$ Dis the midpoint ofBC.So, the coordinates ofDare $D\left(\frac{6+1}{2}, \frac{5+4}{2}\right) \quad\left[B(6,5)\right.$ and $C(1,4) \Rightarrow\left(x_{1}=6, y_{1}=5\right)$ and $\left.\left(x_{2}=1, y_{2}=4\ri...
Read More →How much electricity is required in coulomb for the oxidation of
Question: How much electricity is required in coulomb for the oxidation of (i)1 mol of H2O to O2. (ii)1 mol of FeO to Fe2O3. Solution: (i)According to the question, $\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}$ Now, we can write: $\mathrm{O}^{2-} \longrightarrow \frac{1}{2} \mathrm{O}_{2}+2 \mathrm{e}^{-}$ Electricity required for the oxidation of 1 mol of H2O to O2= 2 F = 2 96487 C = 192974 C (ii)According to the question, $\mathrm{Fe}^{2+} \longrightarro...
Read More →Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm.
Question: Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle, such that the area enclosed between it and the 7 cm circle is same as that between the two inner circles. Find the radius of the third circle correct to one decimal place. Solution: We know the length of OA and OB. $O A=r_{1}=3.5 \mathrm{~cm}$ $O B=r_{2}=7 \mathrm{~cm}$ Now we will calculate the area between the circles of radii 3.5 cm and 7 cm as ...
Read More →The line 2x + y −4 = 0 divides the line segment joining
Question: The line 2x+y4 = 0 divides the line segment joiningA(2, 2) andB(3, 7) in the ratio(a) 2 : 5(b) 2 : 9(c) 2 : 7(d) 2 : 3 Solution: (b) 2 : 9 Let the line $2 x+y-4=0$ divide the line segment in the ratio $k: 1$ at the point $P$. Then, by section formula, the coordinatesofPare $P\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right)$ Since $P$ lies on the line $2 x+y-4=0$, we have: $\frac{2(3 k+2)}{k+1}+\frac{7 k-2}{k+1}-4=0$ $\Rightarrow(6 k+4)+(7 k-2)-(4 k+4)=0$ $\Rightarrow 9 k=2$ $\Rightarr...
Read More →How much electricity in terms of Faraday is required to produce
Question: How much electricity in terms of Faraday is required to produce (i)20.0 g of Ca from molten CaCl2. (ii)40.0 g of Al from molten Al2O3. Solution: (i)According to the question, Electricity required to produce 40 g of calcium = 2 F Therefore, electricity required to produce $20 \mathrm{~g}$ of calcium $=\frac{2 \times 20}{40} \mathrm{~F}$ = 1 F (ii)According to the question, Electricity required to produce 27 g of Al = 3 F Therefore, electricity required to produce $40 \mathrm{~g}$ of $\m...
Read More →How much charge is required for the following reductions:
Question: How much charge is required for the following reductions: (i)1 mol of Al3+to Al. (ii)1 mol of Cu2+to Cu. (iii) $1 \mathrm{~mol}$ of $\mathrm{MnO}_{4}^{-}$to $\mathrm{Mn}^{2+}$. Solution: (i) $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}$ $\therefore$ Required charge $=3 \mathrm{~F}$ = 3 96487 C = 289461 C (ii) $\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}$ $\therefore$ Required charge $=2 \mathrm{~F}$ = 2 96487 C = 192974 C (iii) $\mathrm{MnO}_{4}^{-} ...
Read More →If P(−1, 1) is the midpoint of the line segment joining
Question: IfP(1, 1) is the midpoint of the line segment joiningA(3,b) andB(1,b+ 4), thenb= ?(a) 1(b) 1(c) 2(d) 0 Solution: (b) 1The given points areA(3,b) andB(1,b+4). Then, $\left(x_{1}=-3, y_{1}=b\right)$ and $\left(x_{2}=1, y_{2}=b+4\right)$ Therefore, $x=\frac{[(-3)+1]}{2}$ $=\frac{-2}{2}$ $=-1$ and $y=\frac{[b+(b+4)]}{2}$ $=\frac{2 b+4}{2}$ $=b+2$ But the midpoint is $P(-1,1)$. Therefore, $b+2=1$ $\Rightarrow b=-1$...
Read More →From a thin metallic piece, in the shape of a trapezium ABCD,
Question: From a thin metallic piece, in the shape of a trapeziumABCD, in whichAB||CDand BCD= 90, a quarter circleBEFCis removed (in the following figure). GivenAB=BC= 3.5 cm andDE= 2 cm, calculate the area of the remaining piece of the metal sheet. Solution: We have given a trapezium. We are asked to find the area of the shaded region. We can find the area of the remaining part that is area of the shaded region as shown below. Area of the shaded region $=$ Area of the trapezium $-$ Area of the ...
Read More →Conductivity of 0.00241 M acetic acid is
Question: Conductivity of $0.00241 \mathrm{M}$ acetic acid is $7.896 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity and if $\Lambda_{m}^{0}$ for acetic acid is $390.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, what is its dissociation constant? Solution: Given,= 7.896 105S m1 c = 0.00241 mol L1 Then, molar conductivity, $\Lambda_{m}=\frac{\kappa}{\mathrm{c}}$ $=\frac{7.896 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}}{0.00241 \mathrm{~mol} \mathrm{~L}^{-1}...
Read More →In what ratio does the y-axis divide the join of P(−4, 2) and Q(8, 3)?
Question: In what ratio does they-axis divide the join ofP(4, 2) andQ(8, 3)? (a) 3 : 1(b) 1 : 3(c) 2 : 1(d) 1 : 2 Solution: (d) 1 : 2 Let $A B$ be divided by the $y$ axis in the ratio $k: 1$ at the point $P$. Then, by section formula, the coordinates ofPare $P\left(\frac{8 k-4}{k+1}, \frac{3 k+2}{k+1}\right)$ But,Plies on they axis; so, its abscissa is 0 $\Rightarrow \frac{8 k-4}{k+1}=0$ $\Rightarrow 8 k-4=0$ $\Rightarrow 8 k=4$ $\Rightarrow k=\frac{1}{2}$ Hence, the required ratio is $\frac{1}{...
Read More →ABCDEF is a regular hexagon with centre O (in the following figure).
Question: ABCDEFis a regular hexagon with centre O (in the following figure). If the area of triangleOABis 9 cm2, find the area of : (i) the hexagon and (ii) the circle in which the haxagon is incribed. Solution: We know that a regular hexagon is made up of 6 equilateral triangles. We have given area of the one of the triangles. $\therefore$ Area of the hexagon $=6 \times$ area of one equilateral triangle $\therefore$ Area of the hexagon $=6 \times 9$ $\therefore$ Area of the hexagon $=54$ We kn...
Read More →The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
Question: The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100 102 /S m11.237 11.85 23.15 55.53 106.74 Calculate $\Lambda_{m}$ for all concentrations and draw a plot between $\Lambda_{m}$ and $\mathrm{c}^{1 / 2}$. Find the value of $\Lambda_{m}^{0}$. Solution: Given, = 1.237 102S m1, c = 0.001 M Then,= 1.237 104S cm1,c = 0.0316 M1/2 $\therefore \Lambda_{m}=\frac{\kappa}{c}$ $=\f...
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