The earth’s surface has a negative surface charge density of
Question: The earth's surface has a negative surface charge density of $10^{-9} \mathrm{C} \mathrm{m}^{-2}$. The potential difference of $400 \mathrm{kV}$ between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth's surface? (This never happens in practice beca...
Read More →Solve the equation 21x2 – 28x + 10 = 0
Question: Solve the equation $21 x^{2}-28 x+10=0$ Solution: The given quadratic equation is $21 x^{2}-28 x+10=0$ On comparing the given equation with $a x^{2}+b x+c=0$, we obtain $a=21, b=-28$, and $c=10$ Therefore, the discriminant of the given equation is $D=b^{2}-4 a c=(-28)^{2}-4 \times 21 \times 10=784-840=-56$ Therefore, the required solutions are $\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-28) \pm \sqrt{-56}}{2 \times 21}=\frac{28 \pm \sqrt{56} i}{42}$ $=\frac{28 \pm 2 \sqrt{14} i}{42}=\frac{28...
Read More →The number density of free electrons in a copper conductor estimated in Example
Question: The number density of free electrons in a copper conductor estimated in Example $3.1$ is $8.5 \times 10^{28} \mathrm{~m}^{-3}$. How long does an electron take to drift from one end of a wire $3.0 \mathrm{~m}$ long to its other end? The area of cross-section of the wire is $2.0 \times 10^{-6} \mathrm{~m}^{2}$ and it is carrying a current of $3.0 \mathrm{~A}$. Solution: Number density of free electrons in a copper conductor,n= 8.5 1028m3Length of the copper wire,l= 3.0 m Area of cross-se...
Read More →Whenever a reaction between an oxidising agent and a reducing agent is carried out,
Question: Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations. Solution: Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compo...
Read More →Prove
Question: Prove $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}$ Solution: Let $\sin ^{-1} \frac{8}{17}=x$. Then, $\sin x=\frac{8}{17} \Rightarrow \cos x=\sqrt{1-\left(\frac{8}{17}\right)^{2}}=\sqrt{\frac{225}{289}}=\frac{15}{17}$. $\therefore \tan x=\frac{8}{15} \Rightarrow x=\tan ^{-1} \frac{8}{15}$ $\therefore \sin ^{-1} \frac{8}{17}=\tan ^{-1} \frac{8}{15}$ ....(i) Now, let $\sin ^{-1} \frac{3}{5}=y$. Then, $\sin y=\frac{3}{5} \Rightarrow \cos y=\sqrt{1-\left(\frac{3...
Read More →Express the following rational numbers as decimals
Question: Express the following rational numbers as decimals: (i) $\frac{2}{3}$ (ii) $-\frac{4}{9}$ (iii) $\frac{-2}{15}$ (iv) $-\frac{22}{13}$ (v) $\frac{437}{999}$ (vi) $\frac{33}{26}$ Solution: (i) Given rational number is Now we have to express this rational number in decimal form. So we will use the long division method Therefore $\frac{2}{3}=0.6666$ $\Rightarrow \frac{2}{3}=0 . \overline{6}$ Hence, $\frac{2}{3}=0 . \overline{6}$ (ii) Given rational number is Now we have to express this rat...
Read More →In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire.
Question: In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell? Solution: Emf of the cell,E1= 1.25 V Balance point of the potentiometer,l1= 35 cm The cell is replaced by another cell of emfE2. New balance point of the potentiometer,l2= 63 cm The balance condition is given by the relation, $\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$...
Read More →Solve the equation 27x2 – 10x + 1 = 0
Question: Solve the equation $27 x^{2}-10 x+1=0$ Solution: The given quadratic equation is $27 x^{2}-10 x+1=0$ On comparing the given equation with $a x^{2}+b x+c=0$, we obtain $a=27, b=-10$, and $c=1$ Therefore, the discriminant of the given equation is $D=b^{2}-4 a c=(-10)^{2}-4 \times 27 \times 1=100-108=-8$ Therefore, the required solutions are $\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{-(-10) \pm \sqrt{-8}}{2 \times 27}=\frac{10 \pm 2 \sqrt{2} i}{54}$ $[\sqrt{-1}=i]$ $=\frac{5 \pm \sqrt{2}...
Read More →A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω.
Question: A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit? Solution: Emf of the storage battery,E= 8.0 V Internal resistance of the battery,r= 0.5 Ω DC supply voltage,V= 120 V Resistance of the resistor,R= 15.5 Ω Effective voltage in the circuit =V1 Ris connected to the storage ba...
Read More →Find the value of
Question: Find the value of $\cos ^{-1}\left(\cos \frac{15 \pi}{6}\right)$ Solution: We know that $\cos ^{-1}(\cos x)=x$ if $x \in[0, \pi]$, which is the principal value branch of $\cos ^{-1} x$. Here, $\frac{13 \pi}{6} \notin[0, \pi]$. Now, $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ can be written as: $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right]=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]$, where $\frac{\pi}{6} \in[0, ...
Read More →Solve the equation
Question: Solve the equation $x^{2}-2 x+\frac{3}{2}=0$ Solution: The given quadratic equation is $x^{2}-2 x+\frac{3}{2}=0$ This equation can also be written as $2 x^{2}-4 x+3=0$ On comparing this equation with $a x^{2}+b x+c=0$, we obtain $a=2, b=-4$, and $c=3$ Therefore, the discriminant of the given equation is $D=b^{2}-4 a c=(-4)^{2}-4 \times 2 \times 3=16-24=-8$ Therefore, the required solutions are $\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-4) \pm \sqrt{-8}}{2 \times 2}=\frac{4 \pm 2 \sqrt{2} i}...
Read More →The compound AgF2 is an unstable compound. However, if formed,
Question: The compound AgF2is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why? Solution: The oxidation state of $\mathrm{Ag}$ in $\mathrm{AgF}_{2}$ is $+2$. But, $+2$ is an unstable oxidation state of $\mathrm{Ag}$. Therefore, whenever $\mathrm{AgF}_{2}$ is formed, silver readily accepts an electron to form $\mathrm{Ag}^{+}$. This helps to bring the oxidation state of $\mathrm{Ag}$ down from $+2$ to a more stable state of $+1$. As a result, $\mat...
Read More →(a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X.
Question: (a)In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the endA, when the resistorYis of 12.5 Ω. Determine the resistance ofX. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips? (b)Determine the balance point of the bridge above ifXandYare interchanged. (c)What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current? Solution: A metre...
Read More →Consider the reactions:
Question: Consider the reactions: (a) $6 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{~g})$ (b) $\mathrm{O}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+2 \mathrm{O}_{2}(\mathrm{~g})$ Why it is more appropriate to write these reactions as: (a) $6 \mathrm{CO}_{2}(\mathrm{~g})+12 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})...
Read More →Solve the equation
Question: Solve the equation $3 x^{2}-4 x+\frac{20}{3}=0$ Solution: The given quadratic equation is $3 x^{2}-4 x+\frac{20}{3}=0$ This equation can also be written as $9 x^{2}-12 x+20=0$ On comparing this equation with $a x^{2}+b x+c=0$, we obtain $a=9, b=-12$, and $c=20$ Therefore, the discriminant of the given equation is $D=b^{2}-4 a c=(-12)^{2}-4 \times 9 \times 20=144-720=-576$ Therefore, the required solutions are $\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{-(-12) \pm \sqrt{-576}}{2 \times ...
Read More →Determine the current in each branch of the network shown in fig 3.30:
Question: Determine the current in each branch of the network shown in fig 3.30: Solution: Current flowing through various branches of the circuit is represented in the given figure. I1= Current flowing through the outer circuit I2= Current flowing through branch AB I3= Current flowing through branch AD I2I4= Current flowing through branch BC I3+I4= Current flowing through branch CD I4= Current flowing through branch BD For the closed circuit ABDA, potential is zero i.e., $10 l_{2}+5 l_{4}-5 l_{...
Read More →Convert the following in the polar form:
Question: Convert the following in the polar form: (i) $\frac{1+7 i}{(2-i)^{2}}$ (ii) $\frac{1+3 i}{1-2 i}$ Solution: (i) Here, $z=\frac{1+7 i}{(2-i)^{2}}$ $=\frac{1+7 i}{(2-i)^{2}}=\frac{1+7 i}{4+i^{2}-4 i}=\frac{1+7 i}{4-1-4 i}$ $=\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}=\frac{3+4 i+21 i+28 i^{2}}{3^{2}+4^{2}}$ $=\frac{3+4 i+21 i-28}{3^{2}+4^{2}}=\frac{-25+25 i}{25}$ $=-1+i$ Let $r \cos \theta=-1$ and $r \sin \theta=1$ On squaring and adding, we obtain $r^{2}\left(\cos ^{2} \theta+\sin ^...
Read More →Prove
Question: Prove $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$ Solution: Let $\sin ^{-1} \frac{3}{5}=x$. Then, $\sin x=\frac{3}{5}$. $\Rightarrow \cos x=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5}$ $\therefore \tan x=\frac{3}{4}$ $\therefore x=\tan ^{-1} \frac{3}{4} \Rightarrow \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$ Now, we have: L.H.S. $=2 \sin ^{-1} \frac{3}{5}=2 \tan ^{-1} \frac{3}{4}$ $=\tan ^{-1}\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)$ $\le...
Read More →Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\sin x=\frac{1}{4}, x$ in quadrant II
[question] Question. Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\sin x=\frac{1}{4}, x$ in quadrant II [/question] [solution] solution: Here, x is in quadrant II. $\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$ Therefore, $\sin \frac{x}{2}, \cos \frac{x}{2}$, and $\tan \frac{x}{2}$ are all positive. It is given that $\sin x=\frac{1}{4}$. $\cos ^{2} x=1-\sin ^{2} x=1-\left(\frac{1}{4}\right)^{2}=1-\frac{1}{16}=\frac{15}{16}$ $\Rightarrow \cos x=-\frac{\sqrt{15}}{4}...
Read More →While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions,
Question: While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why? Solution: In sulphur dioxide (SO2), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to 2. Therefore, SO2can act as an oxidising as well as a reducing agent. In hydrogen peroxide (H2O2), the O.N. of O is 1 and the range of the O.N. that O can have is from 0 to 2. O can sometimes also a...
Read More →Suggest a list of the substances where carbon can exhibit oxidation states from
Question: Suggest a list of the substances where carbon can exhibit oxidation states from 4 to +4 and nitrogen from 3 to +5. Solution: The substances where carbon can exhibit oxidation states from 4 to +4 are listedin the following table. The substances where nitrogen can exhibit oxidation states from 3 to +5 are listed in the following table....
Read More →Find the value of
Question: Find the value of $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$ Solution: We know that $\tan ^{-1}(\tan x)=x$ if $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, which is the principal value branch of $\tan ^{-1} x$. Here, $\frac{7 \pi}{6} \notin\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ Now, $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$ can be written as: $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\tan ^{-1}\left[\tan \left(2 \pi-\frac{5 \pi}{6}\right)\right]$ $[\tan (2 \pi-x)=-\...
Read More →Aheating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds toa steady value of 2.8 A.
Question: Aheating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds toa steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 104C1. Solution: Supply voltage,V= 230 V Initial current drawn,I1= 3.2 A Initial resistance =R1, which is given by the relation, $R_{1}=\frac{V...
Read More →Find the value of
Question: Find the value of Solution: We know that $\cos ^{-1}(\cos x)=x$ if $x \in[0, \pi]$, which is the principal value branch of $\cos ^{-1} x$. Here, $\frac{13 \pi}{6} \notin[0, \pi]$. Now, $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ can be written as: $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right]=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]$, where $\frac{\pi}{6} \in[0, \pi]$ $\therefore \cos ^{-1}\left(\cos \frac{13...
Read More →If x – iy =prove that.
Question: If $x-i y=\sqrt{\frac{a-i b}{c-i d}}$ prove that $\left(x^{2}+y^{2}\right)^{2}=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$ Solution: $x-i y=\sqrt{\frac{a-i b}{c-i d}}$ $=\sqrt{\frac{\mathrm{a}-\mathrm{ib}}{\mathrm{c}-\mathrm{id}} \times \frac{\mathrm{c}+\mathrm{id}}{\mathrm{c}+\mathrm{id}}}[$ On multiplying numerator and deno min ator by $(\mathrm{c}+\mathrm{id})]$ $=\sqrt{\frac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}}$ $\therefore(x-i y)^{2}=\frac{(a c+b d)+i(a d-b c)}{c^{2}+d^{2}}$ $\Rightarrow x^{2...
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