If A.M. and G.M. of roots of a quadratic equation are 8 and 5,

Question: If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation. Solution: Let the root of the quadratic equation beaandb. According to the given condition, A.M. $=\frac{a+b}{2}=8 \Rightarrow a+b=16$$\ldots(1)$ G.M. $=\sqrt{a b}=5 \Rightarrow a b=25$ $\ldots(2)$ G.M. $=\sqrt{a b}=5 \Rightarrow a b=25$ The quadratic equation is given by, $x^{2}-x$ (Sum of roots) $+$ (Product of roots) $=0$ $x^{2}-x(a+b)+(a b)=0$ $x^{2}-16 x+25=0[U \operato...

Read More →

(a) Estimate the speed with which electrons emitted from a heated emitter

Question: (a)Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. Thespecific chargeof the electron, i.e., itse/mis given to be 1.76 1011C kg1. (b)Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modif...

Read More →

Why is it necessary to use acetic acid and not sulphuric acid

Question: Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test? Solution: Although the addition of sulphuric acid will precipitate lead sulphate, the addition of acetic acid will ensure a complete precipitation of sulphur in the form of lead sulphate due to common ion effect. Hence, it is necessary to use acetic acid for acidification of sodium extract for testing sulphur by lead acetate test....

Read More →

What will Rs 500 amounts to in 10 years after its deposit in a bank

Question: What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of $10 \%$ compounded annually? Solution: The amount deposited in the bank is Rs 500. At the end of first year, amount $=\operatorname{Rs} 500\left(1+\frac{1}{10}\right)=\operatorname{Rs} 500$ At the end of $2^{\text {nd }}$ year, amount = Rs $500(1.1)(1.1)$ At the end of $3^{\text {rd }}$ year, amount $=$ Rs 500 (1.1) (1.1) (1.1) and so on $\therefore$ Amount at the end of 10 years $=\o...

Read More →

Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during

Question: Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound? Solution: Carbon dioxide is acidic in nature and potassium hydroxide is a strong base. Hence, carbon dioxide reacts with potassium hydroxide to form potassium carbonate and water as $2 \mathrm{KOH}+\mathrm{CO}_{2} \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}$ Thus, the mass of the U-tube containing KOH increases. This...

Read More →

Examine the consistency of the system of equations.

Question: Examine the consistency of the system of equations. $3 x-y-2 z=2$ $2 y-z=-1$ $3 x-5 y=3$ Solution: The given system of equations is: $3 x-y-2 z=2$ $2 y-z=-1$ $3 x-5 y=3$ This system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{ccc}3 -1 -2 \\ 0 2 -1 \\ 3 -5 0\end{array}\right], X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}2 \\ -1 \\ 3\end{array}\right]$. Now, $|A|=3(0-5)-0+3(1+4)=-15+15=0$ $\therefore A$ is a singula...

Read More →

The number of bacteria in a certain culture doubles every hour.

Question: The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally,how many bacteria will be present at the endof $2^{\text {nd }}$ hour, $4^{\text {th }}$ hour and $n^{\text {th }}$ hour? Solution: It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P. Here, $a=30$ and $r=2$ $\therefore a_{3}=a r^{2}=(30)(2)^{2}=120$ Therefore, the number of bacteria at the...

Read More →

Will CCl4 give white precipitate of AgCl on heating it with silver nitrate?

Question: Will $\mathrm{CCl}_{4}$ give white precipitate of $\mathrm{AgCl}$ on heating it with silver nitrate? Give reason for your answer. Solution: $\mathrm{CCl}_{4}$ will not give the white precipitate of $\mathrm{AgCl}$ on heating it with silver nitrate. This is because the chlorine atoms are covalently bonded to carbon in $\mathrm{CCl}_{4}$. To obtain the precipitate, it should be present in ionic form and for this, it is necessary to prepare the Lassaigne's extract of $\mathrm{CCl}_{4}$....

Read More →

Will CCl4 give white precipitate of AgCl on heating it with silver nitrate?

Question: Will $\mathrm{CCl}_{4}$ give white precipitate of $\mathrm{AgCl}$ on heating it with silver nitrate? Give reason for your answer. Solution: $\mathrm{CCl}_{4}$ will not give the white precipitate of $\mathrm{AgCl}$ on heating it with silver nitrate. This is because the chlorine atoms are covalently bonded to carbon in $\mathrm{CCl}_{4}$. To obtain the precipitate, it should be present in ionic form and for this, it is necessary to prepare the Lassaigne's extract of $\mathrm{CCl}_{4}$....

Read More →

What is the de Broglie wavelength of a nitrogen molecule in air at 300 K?

Question: What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u) Solution: Temperature of the nitrogen molecule,T= 300 K Atomic mass of nitrogen = 14.0076 u Hence, mass of the nitrogen molecule,m= 2 14.0076 = 28.0152 u But 1 u = 1.66 1027kg m= 28.0152 1.66 1027kg Plancks constant,h= 6.63 1034Js Boltzmann constant, $k=1.38 \times 10^{-23} \m...

Read More →

What is the de Broglie wavelength of a nitrogen molecule in air at 300 K?

Question: What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u) Solution: Temperature of the nitrogen molecule,T= 300 K Atomic mass of nitrogen = 14.0076 u Hence, mass of the nitrogen molecule,m= 2 14.0076 = 28.0152 u But 1 u = 1.66 1027kg m= 28.0152 1.66 1027kg Plancks constant,h= 6.63 1034Js Boltzmann constant, $k=1.38 \times 10^{-23} \m...

Read More →

Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?

Question: Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation? Solution: In steam distillation, the organic liquid starts to boil when the sum of vapour pressure due to the organic liquid $\left(p_{1}\right)$ and the vapour pressure due to water $\left(p_{2}\right)$ becomes equal to atmospheric pressure $(p)$, that is, $p=p_{1}+p_{2}$ Since $p_{1}p_{2}$, organic liquid will vapourise at a lower temperature than its boiling point....

Read More →

If A and G be A.M. and G.M., respectively between two positive numbers,

Question: If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are$\mathrm{A} \pm \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}$ Solution: It is given thatAandGare A.M. and G.M. between two positive numbers. Let these two positive numbers beaandb. $\therefore \mathrm{AM}=\mathrm{A}=\frac{\mathrm{a}+\mathrm{b}}{2}$$\ldots(1)$ $\mathrm{GM}=\mathrm{G}=\sqrt{\mathrm{ab}}$ ..(2) From $(1)$ and $(2)$, we obtain $a+b=2 A \ldots$ (3) $a b=G^{2} \ldots(4)...

Read More →

Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor

Question: Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor Solution: The process of sublimation is used to separate a mixture of camphor and calcium sulphate. In this process, the sublimable compound changes from solid to vapour state without passing through the liquid state. Camphor is a sublimable compound and calcium sulphate is a non-sublimable solid. Hence, on heating, camphor will sublime while calcium sulphate will be left behind....

Read More →

Examine the consistency of the system of equations.

Question: Examine the consistency of the system of equations. $x+y+z=1$ $2 x+3 y+2 z=2$ $a x+a y+2 a z=4$ Solution: The given system of equations is: $x+y+z=1$ $2 x+3 y+2 z=2$ $a x+a y+2 a z=4$ This system of equations can be written in the formAX=B, where $A=\left[\begin{array}{lll}1 1 1 \\ 2 3 2 \\ a a 2 a\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 2 \\ 4\end{array}\right]$. Now $\begin{aligned}|A| =1(6 a-2 a)-1(4 a-2 a)+1(2 a-3 ...

Read More →

Show that the wavelength of electromagnetic radiation is equal

Question: Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon). Solution: The momentum of a photon having energy (h) is given as: p=\frac{h v}{c}=\frac{h}{\lambda} $\lambda=\frac{h}{p}$ ...(i) Where, = Wavelength of the electromagnetic radiation c= Speed of light h= Plancks constant De Broglie wavelength of the photon is given as: $\lambda=\frac{h}{m v}$ But $p=m v$ $\therefore \lambda=\frac{h}{p}$ ...(ii) Where, m= Mass of the photo...

Read More →

Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen,

Question: Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens. Solution: Nitrogen, sulphur, and halogens are covalently bonded in organic compounds. For their detection, they have to be first converted to ionic form. This is done by fusing the organic compound with sodium metal. This is called Lassaignes test. The chemical equations involved in the test are $\mathrm{Na}+\mathrm{C}+\mathrm{N} \longrightarrow \mathrm{NaCN}$ $\mat...

Read More →

(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be

Question: (a)For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 1010m? (b)Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2)kTat 300 K. Solution: (a)De Broglie wavelength of the neutron,= 1.40 1010m Mass of a neutron,mn= 1.66 1027kg Plancks constant,h= 6.6 1034Js Kinetic energy (K) and velocity (v) are related as: $K=\frac{1}{2} m_{n} v^{2} \ldots$ (1) De Broglie wavelength () and vel...

Read More →

Examine the consistency of the system of equations.

Question: Examine the consistency of the system of equations. $x+y+z=1$ $2 x+3 y+2 z=2$ $a x+a y+2 a z=4$ Solution: The given system of equations is: $x+y+z=1$ $2 x+3 y+2 z=2$ $a x+a y+2 a z=4$ This system of equations can be written in the formAX=B, where $A=\left[\begin{array}{lll}1 1 1 \\ 2 3 2 \\ a a 2 a\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}1 \\ 2 \\ 4\end{array}\right]$. Now $\begin{aligned}|A| =1(6 a-2 a)-1(4 a-2 a)+1(2 a-3 ...

Read More →

Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

Question: Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens? Solution: While testing the Lassaignes extract for the presence of halogens, it is first boiled with dilute nitric acid. This is done to decompose NaCN to HCN and Na2S to H2S and to expel these gases. That is, if any nitrogen and sulphur are present in the form of NaCN and Na2S, then they are removed. The chemical equations involved in the reaction are represented as $\mathrm{NaCN}+\mathrm{HNO...

Read More →

The sum of two numbers is 6 times their geometric mean,

Question: The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio$(3+2 \sqrt{2}):(3-2 \sqrt{2})$ Solution: Let the two numbers beaandb. G.M. $=\sqrt{a b}$ According to the given condition, $a+b=6 \sqrt{a b}$$\ldots(1)$ $\Rightarrow(a+b)^{2}=36(a b)$ $\mathrm{Also}$ $(a-b)^{2}=(a+b)^{2}-4 a b=36 a b-4 a b=32 a b$ $\Rightarrow a-b=\sqrt{32} \sqrt{a b}$ $=4 \sqrt{2} \sqrt{a b}$ $\ldots(2)$ Adding (1) and (2), we obtain $2 a=(6+4 \sqrt{2}) \sqrt{a b}$ $\Rightarrow ...

Read More →

Explain the principle of paper chromatography.

Question: Explain the principle of paper chromatography. Solution: In paper chromatography, chromatography paper is used. This paper contains water trapped in it, which acts as the stationary phase. On the base of this chromatography paper, the solution of the mixture is spotted. The paper strip is then suspended in a suitable solvent, which acts as the mobile phase. This solvent rises up the chromatography paper by capillary action and in the procedure, it flows over the spot. The components ar...

Read More →

Examine the consistency of the system of equations.

Question: Examine the consistency of the system of equations. x+ 3y= 5 2x+ 6y= 8 Solution: The given system of equations is: x+ 3y= 5 2x+ 6y= 8 The given system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{ll}1 3 \\ 2 6\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}5 \\ 8\end{array}\right]$ Now, $|A|=1(6)-3(2)=6-6=0$ $\therefore A$ is a singular matrix. Now, $(\operatorname{adj} A)=\left[\begin{array}{cc}6 -3 \\ ...

Read More →

An electron and a photon each have a wavelength of 1.00 nm. Find

Question: An electron and a photon each have a wavelength of 1.00 nm. Find (a)their momenta, (b)the energy of the photon, and (c)the kinetic energy of electron. Solution: Wavelength of an electron $\left(\lambda_{e}\right)$ and a photon $\left(\lambda_{p}\right), \lambda_{e}=\lambda_{p}=\lambda=1 \mathrm{~nm}$ = 1 109m Plancks constant,h= 6.63 1034Js (a)The momentum of an elementary particle is given by de Broglie relation: $\lambda=\frac{h}{p}$ $p=\frac{h}{\lambda}$ It is clear that momentum de...

Read More →

Discuss the principle of estimation of halogens,

Question: Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound. Solution: Estimation of halogens Halogens are estimated by the Carius method. In this method, a known quantity of organic compound is heated with fuming nitric acid in the presence of silver nitrate, contained in a hard glass tube called the Carius tube, taken in a furnace. Carbon and hydrogen that are present in the compound are oxidized to form CO2and H2O respectively and the halog...

Read More →