A bucket is raised from a well by means of
Question: A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm (Fig .15.78). Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/s. Calculate the number of complete revolutions the wheel makes in raising the bucket. Solution: A bucket is pulled from a well using a rope. We have, Velocity by which rope is pulled $(v)=1.1 \mathrm{~m} / \mathrm{s}$ Diameter of the pulley $(d)=77 \mathrm{~cm}$ Time taken to pull the rope, $(t...
Read More →If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (ā2, 5),
Question: If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (2, 5), then the coordinates of the other end of the diameter are (a) $(-6,7)$ (b) $(6,-7)$ (c) $(4,2)$ (d) $(5,3)$ Solution: Let (x,y) be the coordinates of the other end of the diameter. Then $-2=\frac{2+x}{2} \Rightarrow x=-6$ $5=\frac{3+y}{2} \Rightarrow y=7$ Hence, the correct answer is option (a)....
Read More →Given the standard electrode potentials,
Question: Given the standard electrode potentials, K+/K = 2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = 2.37 V, Cr3+/Cr = 0.74V Arrange these metals in their increasing order of reducing power. Solution: The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K+/K Mg2+/Mg Cr3+/Cr Hg2+/Hg Ag+/Ag. Hence, the reducing power of the given metals increases in the following order: Ag Hg Cr Mg K...
Read More →Arrange the following metals in the order in which they displace each other from the solution of their salts.
Question: Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn Solution: The following is the order in which the given metals displace each other from the solution of their salts. Mg, Al, Zn, Fe, Cu...
Read More →The coordinates of the point P dividing the line segment joining the points
Question: The coordinates of the pointPdividing the line segment joining the pointsA(1, 3) andB(4, 6) in the ratio 2 : 1 is (a) $(2,4)$ (b) $(3,5)$ (c) $(4,2)$ (d) $(5,3)$ Solution: Here, the pointPdivides the line segment joining the pointsA(1, 3) andB(4, 6) in the ratio 2 : 1. Then Coordinates of $P=\left(\frac{2 \times 4+1 \times 1}{2+1}, \frac{2 \times 6+1 \times 3}{2+1}\right)$ $=\left(\frac{8+1}{3}, \frac{12+3}{3}\right)$ $=\left(\frac{9}{3}, \frac{15}{3}\right)$ $=(3,5)$ Hence, the correc...
Read More →In the following figure,
Question: In the following figure, PQRS is a square of side 4 cm. Find the area of the shaded square. Solution: Area of the shaded region is equal to the area of the square minus area of four sectors with the same radius minus area of the circle. We know that area of the four sectors with radius is equal to area of one circle. $\therefore$ Area of the shaded region $=4^{2}-\pi \times 1^{2}-\pi \times 1^{2}$ $\therefore$ Area of the shaded region $=4^{2}-2 \times \pi \times 1$ $\therefore$ Area o...
Read More →Explain how rusting of iron is envisaged as setting up
Question: Explain how rusting of iron is envisaged as setting up of an electrochemical cell. Solution: In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by, $\mathrm{Fe}_{(s)} \longrightarrow \mathrm{Fe}^{2+}{ }_{(a q)}+2 \mathrm{e}^{-}$ Electrons released at the anodic spot move through the metallic object and go to another spot of the obje...
Read More →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). The length of one of its diagonals is (a) 5 (b) 4 (c) 3 (d) 25 Solution: Here,ACandBDare two diagonals of the rectangleABCD. So $B D=\sqrt{(4-0)^{2}+(0-3)^{2}}$ $=\sqrt{(4)^{2}+(-3)^{2}}$ $=\sqrt{16+9}$ $=\sqrt{25}$ $=5$ units Hence, the correct answer is option (a)....
Read More →Suggest two materials other than
Question: Suggest two materials other than hydrogen that can be used as fuels in fuel cells. Solution: Methane and methanol can be used as fuels in fuel cells....
Read More →Write the chemistry of recharging the lead storage battery,
Question: Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging. Solution: A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide (PbO2) as the cathode, and a 38% solution of sulphuric acid (H2SO4) as an electrolyte. When the battery is in use, the following cell reactions take place: At anode: $\mathrm{Pb}_{(s)}+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow \mathrm{PbSO}_{4(s)}+2 \mathrm{e}^{-}$...
Read More →In Fig. 15.76, O is the centre of circle of radius 28 cm.
Question: In Fig. 15.76, O is the centre of circle of radius 28 cm. Find the area of minor segmentASB. Solution: We have given that radius of the circle is $28 \mathrm{~cm}$ and sector angle is $45^{\circ}$. Now we will find the area of the minor segment ASB. $\therefore$ Area of the minor segment $A S B=$ Area of the sector $O A S B$ - Area of the $\triangle O A B$ $\therefore$ Area of the minor segment $\mathrm{ASB}=\frac{\theta}{360} \times \pi r^{2}-\frac{1}{2} \times O A \times O B$ $\there...
Read More →Solve this
Question: If $P\left(\frac{a}{2}, 4\right)$ is the midpoint of the line segment joining the points $A(-6,5)$ and $B(-2,3)$ then the value of $a$ is (a) $-8$ (b) 3 (c) $-4$ (d) 4 Solution: The point $P\left(\frac{a}{2}, 4\right)$ is the midpoint of the line segment joining the points $A(-6,5)$ and $B(-2,3)$. So $\frac{a}{2}=\frac{-6-2}{2}$ $\Rightarrow \frac{a}{2}=-4$ $\Rightarrow a=-8$ Hence, the correct answer is option (a)....
Read More →What is the quantity of electricity in coulombs needed to reduce 1 mol of
Question: What is the quantity of electricity in coulombs needed to reduce 1 mol of$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} ?$ Consider the reaction:$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+8 \mathrm{H}_{2} \mathrm{O}$ Solution: The given reaction is as follows: $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}$ Therefore, to reduce 1 mole of $\mathrm{Cr}_{2}...
Read More →In the following figure, OE = 20 cm. In sector OSFT,
Question: In the following figure,OE= 20 cm. In sectorOSFT, squareOEFGis inscribed. Find the area of the shaded region. Solution: We have to find the area of the shaded portion. We have, $\mathrm{OE}=20 \mathrm{~cm}$ and $\mathrm{OEFG}$ is a square. Use Pythagoras theorem to find OF as, $\mathrm{OF}=\left(\sqrt{(20)^{2}+(20)^{2}}\right) \mathrm{cm}$ $=20 \sqrt{2} \mathrm{~cm}$ Therefore area of the shaded region, Area of the shaded region $=($ Area of quadrant of cicle $)-($ Area of square $)$ S...
Read More →Suggest a list of metals that are extracted electrolytically.
Question: Suggest a list of metals that are extracted electrolytically. Solution: Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically....
Read More →If a current of 0.5 ampere flows through a metallic wire for 2 hours,
Question: If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire? Solution: I= 0.5 A t= 2 hours = 2 60 60 s = 7200 s Thus,Q=It = 0.5 A 7200 s = 3600 C We know thatnumber of electrons. Then, Hence,number of electrons will flow through the wire....
Read More →If a current of 0.5 ampere flows through a metallic wire for 2 hours,
Question: If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire? Solution: I= 0.5 A t= 2 hours = 2 60 60 s = 7200 s Thus,Q=It = 0.5 A 7200 s = 3600 C We know thatnumber of electrons. Then, Then, Hence,number of electrons will flow through the wire....
Read More →Solve this
Question: The area of $\triangle A B O$ with vertices $A(a, 0), O(0,0)$ and $B(0, b)$ in square units is (a) $a b$ (b) $\frac{1}{2} a b$ (c) $\frac{1}{2} a^{2} b^{2}$ (d) $\frac{1}{2} b^{2}$ Solution: LetA(x1= a,y1= 0),O(x2= 0,y2= 0) andB(x3= 0,y3=b) be the given vertices. So Area $(\Delta A B O)=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$ $=\frac{1}{2}|a(0-b)+0(b-0)+0(0-0)|$ $=\frac{1}{2}|-a b|=$ $=\frac{1}{2} a b$ Hence, th...
Read More →The molar conductivity of 0.025 mol Lā1 methanoic acid is
Question: The molar conductivity of 0.025 mol L1 methanoic acid is 46.1 S cm2 mol1. Calculate its degree of dissociation and dissociation constant. Given (H+) = 349.6 S cm2mol1and (HCOO) = 54.6 S cm2mol Solution: $C=0.025 \mathrm{~mol} \mathrm{~L}^{-1}$ $\Lambda_{m}=46.1 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$ $\lambda^{0}\left(\mathrm{H}^{+}\right)=349.6 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$ $\lambda^{0}\left(\mathrm{HCOO}^{-}\right)=54.6 \mathrm{Scm}^{2} \mathrm{~mol}^{-1}$ $\Lambda_{m}^{0}(\mathr...
Read More →The diameter of a wheel of a bus is 90 cm which makes 315
Question: The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. [Use = 22/7]. Solution: We have to calculate the speed of the bus. The wheel makes 315 revolutions per minute. We have, Diameter of the wheel $(d)=90 \mathrm{~cm}$ So distance travelled in one revolution, $=\pi d$ $=\frac{22}{7}(90) \mathrm{cm}$ $=\frac{1980}{7} \mathrm{~cm}$ So, Speed $=$ (Distance travelled in one revolution) (Number of revolution per minute) ...
Read More →Identify the operation performed by the circuit given below:
Question: Identify the operation performed by the circuit given below: NANDORANDNOTCorrect Option: , 3 Solution: (3) When two inputs of NAND gate is shorted, it behaves like a NOT gate so boolen equation will be $y=\overline{\bar{A}+\bar{B}+\bar{C}}$ $y=A \cdot B \cdot C$ Thus, whole arrangement behaves like a AND gate....
Read More →Suggest a way to determine the
Question: Suggest a way to determine the $\Lambda_{m}^{0}$ value of water. Solution: Applying Kohlrausch's law of independent migration of ions, the $\Lambda_{m}^{0}$ value of water can be determined as follows: $\Lambda_{m\left(\mathrm{H}_{2} \mathrm{O}\right)}^{0}=\lambda_{\mathrm{H}^{+}}^{0}+\lambda_{\mathrm{OH}^{-}}^{0}$ $=\left(\lambda_{\mathrm{H}^{+}}^{0}+\lambda_{\mathrm{Cr}}^{0}\right)+\left(\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{OH}^{-}}^{0}\right)-\left(\lambda_{\mathrm{Na}^{+}...
Read More →The area of a triangle with vertices A(5, 0), B(8, 0) and C(8, 4) in square units is
Question: The area of a triangle with verticesA(5, 0),B(8, 0) andC(8, 4) in square units is (a) 20 (b) 12 (c) 6 (d) 16 Solution: LetA(x1= 5,y1= 0),B(x2= 8,y2= 0) andC(x3= 8,y3= 4) be the vertices of the triangle. Then Area $(\Delta A B C)=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$ $=\frac{1}{2}[5(0-4)+8(4-0)+8(0-0)]$ $=\frac{1}{2}[-20+32+0]$ $=6$ sq. units Hence, the correct answer is option (c)....
Read More →Find the area of the circle in which a square
Question: Find the area of the circle in which a square of area $64 \mathrm{~cm}^{2}$ is inscribed. [Use $\left.\pi=3.14\right]$ Solution: We have given area of the square. $\therefore$ side $^{2}=64$ $\therefore$ side $=8$ Now we will find the diameter of the square. $\therefore$ diagonal $=\sqrt{2} \times$ side $\therefore$ diagonal $=\sqrt{2} \times 8$ $\therefore$ diagonal $=8 \sqrt{2}$ We know that diagonal of the square is same as the diameter of the circle. $\therefore$ diameter $=8 \sqrt...
Read More →Why does the conductivity of a solution decrease with dilution?
Question: Why does the conductivity of a solution decrease with dilution? Solution: The conductivity of a solution is the conductance ofions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution....
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