Solve the following
Question: Two salts $A_{2} X$ and $\mathrm{MX}$ have the same value of solubility product of $4.0 \times 10^{-12}$. The ratio of their molar solubilities i.e. $\frac{S\left(A_{2} X\right)}{S(M X)}=$ ______________/(Round off to the Nearest Integer). Solution: (50) For $A_{2} X$ $\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}_{1}^{3}=4 \times 10^{-12}$ $\mathrm{S}_{1}=10^{-4}$ for $M X$ $\mathrm{MX} \rightarrow \mathrm{M}^{+}+\mathrm{X}^{-}$ $\mathrm{S}_{2} \quad \mathrm{~S}_{2}$ $\mathrm{K}_{\mathrm{sp}...
Read More →The common difference of the A.P
Question: The common difference of the A.P. $b_{1}, b_{2}, \ldots, b_{m}$ is 2 more than the common difference of A.P. $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{n}}$. If $\mathrm{a}_{40}=-159$, $\mathrm{a}_{100}=-399$ and $\mathrm{b}_{100}=\mathrm{a}_{70}$, then $\mathrm{b}_{1}$ is equal to:(1) 81(2) $-127$(3) $-81$(4) 127Correct Option: , 3 Solution: Let common difference of series $a_{1}, a_{2}, a_{3}, \ldots \ldots, a_{n}$ be $d$ $\because a_{40}=a_{1}+39 d=-159 \quad \ldot...
Read More →The sum of the first n terms of an AP is given by Sn = (3n2 – 4n) .
Question: The sum of the firstnterms of an AP is given bySn= (3n2 4n) . Find its (i)nth term, (ii) first term and (iii) common difference. Solution: Given: $S_{n}=3 n^{2}-4 n$ Now, Sum of one term $=S_{1}=3(1)^{2}-4(1)$ $=3-4$ $=-1$ $\Rightarrow a_{1}=a=-1 \quad \ldots$ (1) Hence, the first term is $-1$. Sum of two term $s=S_{2}=3(2)^{2}-4(2)$ $=12-8$ $=4$ $\Rightarrow a_{1}+a_{2}=4 \quad \ldots(2)$ Subtracting (1) from (2), we get $a_{2}=5$ $\Rightarrow a+d=5$ $\Rightarrow d=6 \quad(\because a=...
Read More →Let a, b, c, d and p be any non zero
Question: Let $a, b, c, d$ and $p$ be any non zero distinct real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right)$ $=0$. Then :(1) $a, c, p$ are in A.P.(2) $a, c, p$ are in G.P.(3) $a, b, c, d$ are in G.P.(4) $a, b, c, d$ are in A.P.Correct Option: , 3 Solution: Rearrange given equation, we get $\left(a^{2} p^{2}-2 a b p+b^{2}\right)+\left(b^{2} p^{2}-2 b c p+c^{2}\right)$ $+\left(c^{2} p^{2}-2 c d p+d^{2}\right)=0$ $\Rightarrow(a p-b)^{2}+(...
Read More →An ideal fluid flows (laminar flow) through a pipe of nonuniform diameter.
Question: An ideal fluid flows (laminar flow) through a pipe of nonuniform diameter. The maximum and minimum diameters of the pipes are $6.4 \mathrm{~cm}$ and $4.8 \mathrm{~cm}$, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is:(1) $\frac{9}{16}$(2) $\frac{\sqrt{3}}{2}$(3) $\frac{3}{4}$(4) $\frac{81}{256}$Correct Option: 1 Solution: (1) From the equation of continuity $A_{1} v_{1}=A_{2} v_{2}$ Here, $v_{1}$ and $v_{2}$ are the velocities at two ends of p...
Read More →If the sum of the first 20 terms of the
Question: If the sum of the first 20 terms of the series $\log _{\left(7^{1 / 2}\right)} x+\log _{\left(7^{1 / 3}\right)} x+\log _{\left(7^{1 / 4}\right)} x+\ldots$ is 460 , then $x$ is equal to :(1) $7^{2}$(2) $7^{1 / 2}$(3) $e^{2}$(4) $7^{46 / 21}$Correct Option: 1 Solution: $S=\log _{7} x^{2}+\log _{7} x^{3}+\log _{7} x^{4}+\ldots 20$ terms $\because S=460$ $\Rightarrow \log _{7}\left(x^{2} \cdot x^{3} \cdot x^{4} \cdot \ldots \ldots x^{21}\right)=460$ $\Rightarrow \log _{7} x^{(2+3+4 \ldots ...
Read More →A fluid is flowing through a horizontal pipe of varying cross-section
Question: A fluid is flowing through a horizontal pipe of varying cross-section, with speed $v \mathrm{~ms}^{-1}$ at a point where the pressure is $P$ Pascal. At another point where pressure is $\frac{P}{2}$ Pascal its speed is $\mathrm{V} \mathrm{ms}^{-1}$. If the density of the fluid is $\rho \mathrm{kg} \mathrm{m}^{-3}$ and the flow is streamline, then $\mathrm{V}$ is equal to:(1) $\sqrt{\frac{P}{\rho}+v}$(2) $\sqrt{\frac{2 P}{\rho}+v^{2}}$(3) $\sqrt{\frac{P}{2 \rho}+v^{2}}$(4) $\sqrt{\frac{P...
Read More →The sum of the first n terms of an AP is (3n2 + 6n).
Question: The sum of the firstnterms of an AP is (3n2+ 6n). Find thenth term and the 15th term of this AP. Solution: LetSndenotes the sum of firstnterms of the AP. $\therefore S_{n}=3 n^{2}+6 n$ $\Rightarrow S_{n-1}=3(n-1)^{2}+6(n-1)$ $=3\left(n^{2}-2 n+1\right)+6(n-1)$ $=3 n^{2}-3$ $\therefore n^{\text {th }}$ term of the AP, $a_{n}$ $=S_{n}-S_{n-1}$ $=\left(3 n^{2}+6 n\right)-\left(3 n^{2}-3\right)$ $=6 n+3$ Puttingn= 15, we get $a_{15}=6 \times 15+3=90+3=93$ Hence, thenthterm is (6n+ 3) and 1...
Read More →In an experiment to verify Stokes law,
Question: In an experiment to verify Stokes law, a small spherical ball of radius $r$ and density $\rho$ falls under gravity through a distance $h$ in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of $h$ is proportional to: (ignore viscosity of air)(1) $r^{4}$(2) $r$(3) $r^{3}$(4) $r^{2}$Correct Option: 1 Solution: (1) Using, $v^{2}-u^{2}=2 g h$ $\Rightarrow v^{2}-0^{2}=2 g h \...
Read More →Find the sum of first n terms of an AP whose nth term is (5 − 6n).
Question: Find the sum of firstnterms of an AP whosenth term is (5 6n). Hence, find the sum of its first 20 terms. Solution: Letanbe thenthterm of the AP.an=5 6nPuttingn= 1, we getFirst term,a=a1=5 6 1 = 1Puttingn= 2, we geta2=5 6 2 = 7Letdbe the common difference of the AP. $\therefore d=a_{2}-a_{1}=-7-(-1)=-7+1=-6$ Sum of firstnterms of the AP,Sn $=\frac{n}{2}[2 \times(-1)+(n-1) \times(-6)] \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$ $=\frac{n}{2}(-2-6 n+6)$ $=n(2-3 n)$ $=2 n-3 n^{2}$ ...
Read More →If the sum of the second, third and fourth
Question: If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243 , then the sum of the first 50 terms of this G.P. is:(1) $\frac{1}{26}\left(3^{49}-1\right)$(2) $\frac{1}{26}\left(3^{50}-1\right)$(3) $\frac{2}{13}\left(3^{50}-1\right)$(4) $\frac{1}{13}\left(3^{50}-1\right)$Correct Option: , 2 Solution: Let the first term be ' $a$ ' and common ratio be ' $r$ '. $\because \operatorname{ar}\left(1+r+r^{2}\right)=3$...(...
Read More →A hollow spherical shell at outer radius R floats just submerged under the water surface.
Question: A hollow spherical shell at outer radius $R$ floats just submerged under the water surface. The inner radius of the shell is $r$. If the specific gravity of the shell material is $\frac{27}{8}$ w.r.t water, the value of $r$ is :(1) $\frac{8}{9} R$(2) $\frac{4}{9} R$(3) $\frac{2}{3} R$(4) $\frac{1}{3} R$Correct Option: 1 Solution: (1) In equilibrium, $m g=F_{e}$ $F_{B}=V \rho_{0} g$ and mass $=$ volume $\times$ density $\frac{4}{3} \pi\left(R^{3}-r^{3}\right) \rho_{0} g=\frac{4}{3} \pi ...
Read More →Prove the following
Question: If $3^{2 \sin 2 \alpha-1}, 14$ and $3^{4-2 \sin 2 \alpha}$ are the first three terms of an A.P. for some $\alpha$, then the sixth term of this A.P is:(1) 66(2) 81(3) 65(4) 78Correct Option: 1 Solution: Given that $3^{2 \sin 2 \alpha-1}, 14,3^{4-2 \sin 2 \alpha}$ are in A.P. So, $3^{2 \sin 2 \alpha-1}+3^{4-2 \sin 2 \alpha}=28$ $\Rightarrow \frac{3^{2 \sin 2 \alpha}}{3}+\frac{81}{3^{2 \sin 2 \alpha}}=28$ Let $3^{2 \sin 2 \alpha}=x$ $\Rightarrow \frac{x}{3}+\frac{81}{x}=28$ $\Rightarrow x...
Read More →Prove the following
Question: If $2^{10}+2^{9} \cdot 3^{1}+2^{8} \cdot 3^{2}+\ldots+2 \times 3^{9}+3^{10}=\mathrm{S}-2^{11}$ then $\mathrm{S}$ is equal to:(1) $3^{11}-2^{12}$ (2) $3^{11}$(3) $\frac{3^{11}}{2}+2^{10}$(4) $2 \cdot 3^{11}$Correct Option: , 2 Solution: Given sequence are in G.P. and common ratio $\frac{3}{2}$ $\therefore \frac{2^{10}\left(\left(\frac{3}{2}\right)^{11}-1\right)}{\left(\frac{3}{2}-1\right)}=S-2^{11}$ $\Rightarrow 2^{10} \frac{\left(\frac{3^{11}-2^{11}}{2^{11}}\right)}{\frac{1}{2}}=S-2^{1...
Read More →Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d.
Question: Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $d$. The area of the base of both vessels is $S$ but the height of liquid in one vessel is $x_{1}$ and in the other, $x_{2}$. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is :(1) $g d S\left(x_{2...
Read More →An air bubble of radius
Question: An air bubble of radius $1 \mathrm{~cm}$ in water has an upward acceleration $9.8 \mathrm{~cm} \mathrm{~s}^{-2}$. The density of water is $1 \mathrm{gm}$ $\mathrm{cm}^{-3}$ and water offers negligible drag force on the bubble. The mass of the bubble is $\left(\mathrm{g}=980 \mathrm{~cm} / \mathrm{s}^{2}\right)$ Solution: (3) Given: Radius of air bubble $=1 \mathrm{~cm}$, Upward acceleration of bubble, $a=9.8 \mathrm{~cm} / \mathrm{s}^{2}$, $\rho_{\text {water }}=1 \mathrm{~g} \mathrm{~...
Read More →When a long glass capillary tube of radius
Question: When a long glass capillary tube of radius $0.015 \mathrm{~cm}$ is dipped in a liquid, the liquid rises to a height of $15 \mathrm{~cm}$ within it. If the contact angle between the liquid and glass to close to $0^{\circ}$, the surface tension of the liquid, in millinewton $\mathrm{m}^{-1}$, is $\left[\rho_{\text {(liquid) }}=900 \mathrm{kgm}^{-3}\right.$, $g=10 \mathrm{~ms}^{-2}$ ] (Give answer in closest integer)________ Solution: (101) Given : Radius of capillary tube, $r=0.015 \math...
Read More →Find the sum of each of the following arithmetic series:
Question: Find the sum of each of the following arithmetic series: (i) $7+10 \frac{1}{2}+14+\ldots+84$. (ii) $34+32+30+\ldots+10$. (iii) $(-5)+(-8)+(-11)+\ldots+(-230)$ (iv) $5+(-41)+9+(-39)+13+(-37)+17+\ldots+(-5)+81+(-3)$ Solution: (i) The given arithmetic series is $7+10 \frac{1}{2}+14+\ldots+84$. Here, $a=7, d=10 \frac{1}{2}-7=\frac{21}{2}-7=\frac{21-14}{2}=\frac{7}{2}$ and $I=84$ Let the given series containnterms. Then, $a_{n}=84$ $\Rightarrow 7+(n-1) \times \frac{7}{2}=84 \quad\left[a_{n}...
Read More →Prove the following
Question: Let $a_{1}, a_{2}, \ldots . ., a_{n}$ be a given A.P. whose common difference is an integer and $S_{n}=a_{1}+a_{2}+\ldots . .+a_{n}$. If $a_{1}=1, a_{n}=300$ and $15 \leq n \leq 50$, then the ordered pair $\left(S_{n-4}, a_{n-4}\right)$ is equal to :(1) $(2490,249)$(2) $(2480,249)$(3) $(2480,248)$(4) $(2490,248)$Correct Option: , 4 Solution: Given that $a_{1}=1$ and $a_{n}=300$ and $d \in \mathbf{Z}$ $\therefore 300=1+(n-1) d$ $\Rightarrow d=\frac{299}{(n-1)}=\frac{23 \times 13}{(n-1)}...
Read More →Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their volumes is :
Question: Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes is :(1) $4: 1$(2) $0.8: 1$(3) $8: 1$(4) $2: 1$Correct Option: , 3 Solution: (3) According to question, pressure inside, 1st soap bubble, $\Delta P_{1}=P_{1}-P_{0}=0.01=\frac{4 T}{R_{1}}$ ....(1) And $\Delta P_{2}=P_{2}-P_{0}=0.02=\frac{4 T}{R_{2}}$ ....(2) Dividing, equation (ii) by (i), $\frac{1}{2}=\frac{R_{2}}{R_{1}} \Rightarrow R_{1}=2 R_{2}$ Volume $V=\frac{4}{3} \pi R^{3}$ ...
Read More →The total number of isotopes of hydrogen and number of radioactive isotopes among them,
Question: The total number of isotopes of hydrogen and number of radioactive isotopes among them, respectively, are:3 and 13 and 22 and 12 and 0Correct Option: 1 Solution: There are three isotopes of $\mathrm{H}$ out of which only tritium is radioactive, which emits low energy $\beta^{-}$ particles. Its half life is $12.33$ years....
Read More →The minimum value of $
Question: The minimum value of $2^{\sin x}+2^{\cos x}$ is :(1) $2^{-1+\frac{1}{\sqrt{2}}}$(2) $2^{-1+\sqrt{2}}$(3) $2^{1-\sqrt{2}}$(4) $2^{1-\frac{1}{\sqrt{2}}}$Correct Option: , 4 Solution: $\frac{2^{\sin x}+2^{\cos x}}{2} \geq\left(2^{\sin x+\cos x}\right)^{\frac{1}{2}} \quad(\because \mathrm{AM} \geq \mathrm{GM})$ $\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2 \cdot 2^{\frac{\sin x+\cos x}{2}}$ Since, $-2 \leq \sin x+\cos x \leq \sqrt{2}$ $\therefore$ Minimum value of $2^{\frac{\sin x+\cos x}{2}}=...
Read More →The temporary hardness of water is due to:
Question: The temporary hardness of water is due to:$\mathrm{Na}_{2} \mathrm{SO}_{4}$$\mathrm{NaCl}$$\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$$\mathrm{CaCl}_{2}$Correct Option: , 3 Solution: Unly bicarbonates cause temporary hardness, whereas chlorides and sulphates cause permanent hardness....
Read More →A capillary tube made of glass of radius
Question: A capillary tube made of glass of radius $0.15 \mathrm{~mm}$ is dipped vertically in a beaker filled with methylene iodide (surface tension $=0.05 \mathrm{Nm}^{-1}$, density $=667 \mathrm{~kg} \mathrm{~m}^{-3}$ ) which rises to height $h$ in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opp. sides of the capillary) make an angle of $60^{\circ}$ with one another. Then $h$ is close to $\left(g=10 \mathrm{~ms}^{-2}\right)$.(1) $0.049 \mathrm{~m}$(...
Read More →Prove the following
Question: If $1+\left(1-2^{2} \cdot 1\right)+\left(1-4^{2} \cdot 3\right)+\left(1-6^{2} \cdot 5\right)+\ldots \ldots .+\left(1-20^{2} \cdot 19\right)$ $=\alpha-220 \beta$, then an ordered pair $(\alpha, \beta)$ is equal to :(1) $(10,97)$(2) $(11,103)$(3) $(10,103)$(4) $(11,97)$Correct Option: , 2 Solution: The given series is $1+\left(1-2^{2} \cdot 1\right)+\left(1-4^{2} \cdot 3\right)+\left(1-6^{2} \cdot 5\right)+\ldots\left(1-20^{2} \cdot 19\right)$ $S=1+\sum_{r=1}^{10}\left[1-(2 r)^{2}(2 r-1)...
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