ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3).
Question: ABCDis a rectangle whose three vertices areB(4, 0),C(4, 3) andD(0, 3). Find the length of one of its diagonal. Solution: The given vertices areB(4, 0),C(4, 3) andD(0, 3). Here,BDis one of the diagonals. So $B D=\sqrt{(4-0)^{2}+(0-3)^{2}}$ $=\sqrt{(4)^{2}+(-3)^{2}}$ $=\sqrt{16+9}$ $=\sqrt{25}$ $=5$ Hence, the length of the diagonal is 5 units.....
Read More →Based on solute-solvent interactions,
Question: Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN. Solution: n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in then-octane. The order of increasing polarity is: Cyclohexane CH3CN CH3OH KCl Therefore, the order of increasing solubility is: KCl CH3OH CH3CN Cyclohexane...
Read More →A road which is 7 m wide surrounds a circular park whose circumference is 352 m.
Question: A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road. Solution: It is given that the circumference C of circular track is 352 m. We know that the circumference of circle of radiusris $C=2 \pi r$ Substituting the value of C, $352=2 \times \frac{22}{7} \times r$ $352 \times 7=44 r$ $r=\frac{352 \times 7}{44}$ $r=56 \mathrm{~m}$ Thus, the radius of Circular Park is. Since, 7m wide road surrounds the circular park. Then radius of outer ...
Read More →Based on solute-solvent interactions,
Question: Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN. Solution: n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in then-octane. The order of increasing polarity is: Cyclohexane CH3CN CH3OH KCl Therefore, the order of increasing solubility is: KCl CH3OH CH3CN Cyclohexane...
Read More →For the given circuit, comment on the type of transformer used :
Question: For the given circuit, comment on the type of transformer used : Auxilliary transformerAuto transformerStep-up transformerStep down transformerCorrect Option: , 3 Solution: (3) $V_{S}=\frac{P}{i}=\frac{60}{0.11}=545.45$ $V_{P}=220$ $V_{S}V_{P}$ $\Rightarrow$ Step up transformer...
Read More →Suggest the most important type of intermolecular attractive interaction in the following pairs.
Question: Suggest the most important type of intermolecular attractive interaction in the following pairs. (i)n-hexane and n-octane (ii)I2and CCl4 (iii)NaClO4and water (iv)methanol and acetone (v)acetonitrile (CH3CN) and acetone (C3H6O). Solution: (i)Van der Walls forces of attraction. (ii)Van der Walls forces of attraction. (iii)Ion-diople interaction. (iv)Dipole-dipole interaction. (v)Dipole-dipole interaction....
Read More →Four cows are tethered at four corners of a square plot of side 50 m,
Question: Four cows are tethered at four corners of a square plot of side 50 m, so that they just cannot reach one another. What area will be left ungrazed? (Fig. 15.66) Solution: It is given that four cows are tethered at four corner of square ABCD. We have to find the area of plot that will left ungrazed. Let the side of square isa. a = 25 + 25 m = 50 m Area of square $=a^{2}$ $=50 \times 50$ $=2500 m^{2}$ Area of quadrant inside square $=\frac{1}{4} \pi r^{2}$ $=\frac{1}{4} \times \frac{22}{7...
Read More →At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar.
Question: At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration? Solution: Here, T= 300 K = 1.52 bar R = 0.083 bar L K1mol1 Applying the relation, =CRT $\Rightarrow C=\frac{\pi}{\mathrm{R} T}$ $=\frac{1.52 \mathrm{bar}}{0.083 \mathrm{bar} \mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}$ = 0.061 mol Since the volume of the ...
Read More →If the point A(0, 2) is equidistant from the points
Question: If the pointA(0, 2) is equidistant from the pointsB(3,p) andC(p, 5), findp. Solution: The given points areA(0, 2),B(3,p) andC(p, 5). $A B=A C \Rightarrow A B^{2}=A C^{2}$ $\Rightarrow(3-0)^{2}+(p-2)^{2}=(p-0)^{2}+(5-2)^{2}$ $\Rightarrow 9+p^{2}-4 p+4=p^{2}+9$ $\Rightarrow 4 p=4 \Rightarrow p=1$ Hence,p= 1....
Read More →Two elements A and B form compounds having formula
Question: Two elements A and B form compounds having formula AB2and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2lowers the freezing point by 2.3 Kwhereas 1.0 g of AB4lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kkg mol1. Calculate atomic masses of A and B. Solution: We know that, $M_{2}=\frac{1000 \times w_{2} \times k_{f}}{\Delta T_{f} \times w_{1}}$ Then, $M_{\mathrm{AB}_{2}}=\frac{1000 \times 1 \times 5.1}{2.3 \times 20}$ = 110.87 g mol1 $M_{\mathrm{AB}_{4...
Read More →The value of power dissipated across the zener (diode Vz=15V)
Question: The value of power dissipated across the zener diode $\left(\mathrm{V}_{\mathrm{z}}=15 \mathrm{~V}\right)$ connected in the circuit as shown in the figure is $x \times 10^{-1}$ watt. The value of $x$, to the nearest integer, is Solution: (5) Voltage across $\mathrm{R}_{\mathrm{S}}=22-15=7 \mathrm{~V}$ Current through $\mathrm{R}_{\mathrm{S}}=\mathrm{I}=\frac{7}{35}=\frac{1}{5} \mathrm{~A}$ Current through $90 \Omega=\mathrm{I}_{2}=\frac{15}{90}=\frac{1}{6} \mathrm{~A}$ Current through ...
Read More →Points A(−1, y) and B(5, 7) lie on a circle with centre O(2, −3y).
Question: PointsA(1,y) andB(5, 7) lie on a circle with centreO(2, 3y). Find the values ofy. Solution: The given points areA(1,y),B(5, 7) andO(2, 3y).Here,AOandBOare the radii of the circle. So $A O=B O \Rightarrow A O^{2}=B O^{2}$ $\Rightarrow(2+1)^{2}+(-3 y-y)^{2}=(2-5)^{2}+(-3 y-7)^{2}$ $\Rightarrow 9+(4 y)^{2}=(-3)^{2}+(3 y+7)^{2}$ $\Rightarrow 9+16 y^{2}=9+9 y^{2}+49+42 y$ $\Rightarrow 7 y^{2}-42 y-49=0$ $\Rightarrow y^{2}-6 y-7=0$ $\Rightarrow y^{2}-7 y+y-7=0$ $\Rightarrow y(y-7)+1(y-7)=0$ ...
Read More →Four equal circles, each of radius 5 cm,
Question: Four equal circles, each of radius 5 cm, touch each other as shown in the following figure. Find the area included between them (Take = 3.14). Solution: It is given that four equal circle touches each other as shown in figure. Let the side of square isa. $a=5+5$ $=10 \mathrm{~cm}$ Area of square $=a^{2}$ $=10 \times 10$ $=100 \mathrm{~cm}^{2}$ We know that Area of circle of radius $r=\pi r^{2}$ $=3.14 \times 5 \times 5$ $=78.5 \mathrm{~cm}^{2}$ Area of quadrant inside square $=\frac{1}...
Read More →A 5% solution (by mass) of cane sugar in water has freezing point of 271 K.
Question: A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K. Solution: Here, ΔTf= (273.15 271) K = 2.15 K Molar mass of sugar (C12H22O11) = 12 12 + 22 1 + 11 16 = 342 g mol1 5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 5)g = 95 g of water. Now, number of moles of cane sugar $=\frac{5}{342} \mathrm{~mol}$ = 0.0146 mol Therefore, molal...
Read More →If the area of the triangle with vertices (x, 3), (4, 4) and (3, 5) is 4 square units, find x.
Question: If the area of the triangle with vertices (x, 3), (4, 4) and (3, 5) is 4 square units, findx. Solution: Area of the triangle formed by the vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is $\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$. Now, the given vertices are(x, 3), (4, 4) and (3, 5)and the given area is 4 square units. Therefore, Area of triangle $=\frac{1}{2}|x(4-...
Read More →A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K.
Question: A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298K. Calculate: i. molar mass of the solute ii. vapour pressure of water at 298K. Solution: (i) Let, the molar mass of the solute be $\mathrm{M} \mathrm{g} \mathrm{mol}^{-1}$ Now, the no. of moles of solvent (water), $n_{1}=\frac{90 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mo...
Read More →If a ≠ b ≠ c, prove that (a, a2), (b, b2), (0, 0) will not be collinear.
Question: If a b c, prove that (a, a2), (b, b2), (0, 0) will not be collinear. Solution: Let $\mathrm{A}\left(a, a^{2}\right), \mathrm{B}\left(b, b^{2}\right)$ and $\mathrm{C}(0,0)$ be the coordinates of the given points. We know that the area of triangle having vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is $\left|\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]\right|$ square uni...
Read More →A rectangular piece is 20 m long and 15 m wide.
Question: A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part. Solution: It is given that, the quadrants of radiusrhave been cut from the four corners of a rectangular piece is of lengthand width. We have to find the area of remaining part. We know that, Area of rectangle $=I \times w$ $=20 \times 15$ $=300 \mathrm{~m}^{2}$ Area of quadrant $=\frac{1}{4} \pi r^{2}$ $=\frac{1}{4} \times \frac{22}{7} \ti...
Read More →The outer circumference of a circular race-track is 528 m.
Question: The outer circumference of a circular race-track is 528 m. The track is everywhere 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre (Use = 22/7) Solution: It is given that the outer circumferenceCof circular track is 528 m. We know that the circumference of circle of radiusris $C=2 \pi r$ Substituting the value ofC, $528=2 \times \frac{22}{7} \times r$ $528 \times 7=44 r$ $r=\frac{528 \times 7}{44}$ $r=84 \mathrm{~m}$ Thus, the radius of out...
Read More →Find the area of
Question: Find the area of $\triangle A B C$ with vertices $A(0,-1), B(2,1)$ and $C(0,3)$. Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 : 1. Solution: LetA(x1= 0,y1= 1),B(x2= 2,y2= 1) andC(x3= 0,y3 = 3) be the given points. Then Area $(\Delta A B C)=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$ $=\frac{1}{2}[0(1-3)+2(3+1)+0(-1-1)]$ $=\fr...
Read More →A uniformly thick wheel with moment of inertia
Question: A uniformly thick wheel with moment of inertia I and radius $R$ is free to rotate about its centre of mass (see fig). A massless string is wrapped over its rim and two blocks of masses $m_{1}$ and $m_{2}\left(m_{1}m_{2}\right)$ are attached to the ends of the string. The system is released from rest. The angular speed of the wheel when $m_{1}$ descents by a distance $h$ is: $\left[\frac{2\left(m_{1}-m_{2}\right) g h}{\left(m_{1}+m_{2}\right) \mathrm{R}^{2}+1}\right]^{1 / 2}$$\left[\fra...
Read More →A play ground has the shape of a rectangle,
Question: A play ground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take = 22.7). Solution: It is given that a play ground has a shape of rectangle, with two semicircles on its smaller sides as diameter, added to its outside. So, Area of play ground $=$ Area of rectangle $+2 \times$ Area of semicircle We have, sides of rectanglel= 36 mandb = 24.5 m...
Read More →Calculate the mass of a non-volatile solute
Question: Calculate the mass of a non-volatile solute (molar mass 40 g mol1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%. Solution: Let the vapour pressure of pure octane be $p_{1}^{0}$. Then, the vapour pressure of the octane after dissolving the non-volatile solute is $\frac{80}{100} p_{1}^{0}=0.8 p_{1}^{0}$. Molar mass of solute,M2= 40 g mol1 Mass of octane,w1= 114 g Molar mass of octane, (C8H18),M1= 8 12 + 18 1 = 114 g mol1 Applying the relation, $\frac{p_{...
Read More →One end of a straight uniform 1 m long bar is pivoted on horizontal table.
Question: One end of a straight uniform $1 \mathrm{~m}$ long bar is pivoted on horizontal table. It is released from rest when it makes an angle $30^{\circ}$ from the horizontal (see figure). Its angular speed when it hits the table is given as $\sqrt{n} s^{-1}$, where $n$ is an integer. The value of $n$ is_____ Solution: (15) Here, length of bar, $\mathrm{l}=1 \mathrm{~m}$ angle, $\theta=30^{\circ}$ $\Delta P E=\Delta K E$ or $m g h=\frac{1}{2} I \omega^{2}$ $\Rightarrow(\mathrm{mg}) \frac{l}{2...
Read More →A plot is in the form of a rectangle ABCD having semi-circle on BC
Question: A plot is in the form of a rectangleABCDhaving semi-circle onBCas shown in the following figure. IfAB= 60 m andBC= 28 m, find the area of the plot. Solution: It is given that a plot is in form of rectangle ABCD having a semicircle on BC. $\mathrm{AB}=60 \mathrm{~m}$ $\mathrm{BC}=28 \mathrm{~m}$ Since BC is the diameter of semicircle. Then, radius of semicircle is $r=\frac{28}{2} \mathrm{~m}$ $=14 \mathrm{~m}$ Area of semicircle $=\frac{1}{2} \times \pi r^{2}$ $=\frac{1}{2} \times \frac...
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