If the surface area of a sphere is (144π) m2,
Question: If the surface area of a sphere is (144) m2, then its volume is (a) $(288 \pi) \mathrm{m}^{3}$ (b) $(188 \pi) \mathrm{m}^{3}$ (c) $(300 \pi) \mathrm{m}^{3}$ (d) $(316 \pi) \mathrm{m}^{3}$ Solution: (a) $(288 \pi) \mathrm{m}^{3}$ Surface area $=(144 \pi) \mathrm{m}^{2}$ Lerrm be the radius of the sphere.Then we have: $4 \pi r^{2}=144 \pi$ $\Rightarrow r^{2}=\frac{144}{4}=36$ $\Rightarrow r=6 \mathrm{~m}$ $\therefore$ Volume of the sphere $=\frac{4}{3} \pi r^{3}$ $=\frac{4}{3} \pi \times...
Read More →If the number of integral terms in the expansion of
Question: If the number of integral terms in the expansion of $\left(3^{1 / 2}+5^{1 / 8}\right)^{n}$ is exactly 33 , then the least value of $n$ is :(1) 264(2) 128(3) 256(4) 248Correct Option: , 3 Solution: Here, $\left(3^{2}+5^{8}\right)$ $T_{r+1}={ }^{n} C_{r}(3)^{\frac{n-r}{2}}(5)^{\frac{r}{8}}$ $\because \frac{n-r}{2}$ and $\frac{r}{8}$ are integer So, $r$ must be $0,8,16,24 \ldots \ldots$ Now $n=t_{33}=a+(n-1) d=0+32 \times 8=256$ $\Rightarrow n=256$...
Read More →A certain orbital has n = 4 And
Question: A certain orbital has $\mathrm{n}=4$ and $\mathrm{m}_{\mathrm{L}}=-3$. The number of radial nodes in this orbital is___________ (Round off to the Nearest Integer). Solution: (0) $\mathrm{n}=4$ and $\mathrm{m}_{\ell}=-3$ Hence, $\ell$ value must be 3 . Now, number of radial nodes $=\mathrm{n}-\ell-1$ $=4-3-1=0$...
Read More →For a positive integer
Question: For a positive integer $n,\left(1+\frac{1}{x}\right)^{n}$ is expanded in increasing powers of $x$. If three consecutive coefficients in this expansion are in the ratio, $2: 5: 12$, then $n$ is equal to__________. Solution: According to the question, ${ }^{n} C_{r-1}:{ }^{n} C_{r}:{ }^{n} C_{r+1}=2: 5: 12$ $\Rightarrow \frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{5}{2} \Rightarrow \frac{n-r+1}{r}=\frac{5}{2}$ $\Rightarrow 2 n-7 r+2=0$ ...........(1) $\frac{{ }^{n} C_{r+1}}{{ }^{n} C_{r}}...
Read More →What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25 ,
Question: What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25 , in it's aqueous solution?$5.92$5zero$5.26$Correct Option: 1 Solution: Electronic configuration of divalent metal ion having atomic number 25 is Total number of unpaired electrons $=5$ $\mu($ Magnetic moment $)=\sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{BM}$ where $\mathrm{n}=$ number of unpaired $\mathrm{e}^{-}$ $\therefore \mu=\sqrt{5(5+2)}=\sqrt{35} \mathrm{BM}=5.92 \mathrm{BM}$...
Read More →The surface area of a sphere is 1386 cm2. Its volume is
Question: The surface area of a sphere is 1386 cm2. Its volume is(a) 1617 cm3(b) 3234 cm3(c) 4851 cm3(d) 9702 cm3 Solution: (c) $4851 \mathrm{~cm}^{3}$ Surface area $=1386 \mathrm{~cm}^{2}$ Letrcm be the radius of the sphere.Then we have: $4 \pi r^{2}=1386$ $\Rightarrow r^{2}=\frac{1386 \times 7}{4 \times 22}=110.25$ $\Rightarrow r=10.5 \mathrm{~cm}$ $\therefore$ Volume of the sphere $=\frac{4}{3} \pi r^{3}$ $=\frac{4}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5$ $=4851 \mathrm{~c...
Read More →Let alpha>0, beta>0 be such that $lpha^{3}+eta^{2}=4$
Question: Let $\alpha0, \beta0$ be such that $\alpha^{3}+\beta^{2}=4$. If the maximum value of the term independent of $x$ in the binomial expansion of $\left(\alpha x^{\frac{1}{9}}+\beta x^{-\frac{1}{6}}\right)^{10}$ is $10 k$, then $k$ is equal to :(1) 336(2) 352(3) 84(4) 176Correct Option: 1 Solution: General term of $\left(\alpha x^{\frac{1}{9}}+\beta x^{\frac{-1}{6}}\right)^{10}={ }^{10} C_{r}\left(\alpha x^{\frac{1}{9}}\right)^{10-r}\left(\beta x^{\frac{-1}{6}}\right)^{r}$ $={ }^{10} C_{r}...
Read More →The surface area of a sphere of radius 21 cm is
Question: The surface area of a sphere of radius 21 cm is(a) 2772 cm2(b) 1386 cm2(c) 4158 cm2(d) 5544 cm2 Solution: (d) $5544 \mathrm{~cm}^{2}$ Surface area of sphere $=4 \pi r^{2}$ $=4 \times \frac{22}{7} \times 21 \times 21$ $=5544 \mathrm{~cm}^{2}$...
Read More →The volume of a sphere of radius 10.5 cm is
Question: The volume of a sphere of radius 10.5 cm is(a) 9702 cm3(b) 4851 cm3(c) 19404 cm3(d) 14553 cm3 Solution: (b) 4851 cm3 Volume of the sphere $=\frac{4}{3} \pi r^{3}$ $=\frac{4}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5$ $=88 \times 0.5 \times 10.5 \times 10.5=4851 \mathrm{~cm}^{3}$...
Read More →The number of orbitals with n =5,
Question: The number of orbitals with $\mathrm{n}=5, \mathrm{~m}_{1}=+2$ is ______________ (Round off to the Nearest Integer). Solution: (3) For, $\mathrm{n}=5$ $\ell=(0,1,2,3,4)$ If $\ell=0, \mathrm{~m}=0$ $\ell=1, \mathrm{~m}=\{-1,0,+1\}$ $\ell=2, \mathrm{~m}=\{-2,-1,0,+1,+2\}$ $\ell=3, \mathrm{~m}=\{-3,-2,-1,0,+1,+2,+3\}$ $\ell=4, \mathrm{~m}=\{-4,-3,-2,-1,0,+1,+2,+3,+4\}$ $5 \mathrm{~d}, 5 \mathrm{f}$ and $5 \mathrm{~g}$ subshell contain one-one orbital having $\mathrm{m}_{c}=+2$...
Read More →Consider an electron in a hydrogen atom,
Question: Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65 A$ ). The de-Broglie wavelength of this electron is :(1) $3.5 A$(2) $6.6 A$(3) $12.9 A$(4) $9.7 A$Correct Option: 4, Solution: (4) $v=\frac{c}{137 n}=\frac{c}{137 \times 3}$ $\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{h}{\left(\frac{m \times c}{3 \times 137}\right)}=\frac{h}{m c} \times(3 \times 137)=9.7 A$...
Read More →Let m, n in N$ and gcd (2, n)=1. If
Question: Let $m, n \in N$ and $\operatorname{gcd}(2, n)=1$. If $30\left(\begin{array}{c}30 \\ 0\end{array}\right)+29\left(\begin{array}{c}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)=\mathrm{n} \cdot 2^{\mathrm{m}}$ then $\mathrm{n}+\mathrm{m}$ is equal to Solution: Let $S=\sum_{r=0}^{30}(30-r)^{30} C_{r}$ $=30 \sum_{r=0}^{30}{ }^{30} C_{r}-\sum_{r=0}^{30} r^{30} C_{r}$ $=20 \times 2^{30}-\sum_{r=1}^{30} r \...
Read More →The volume of a sphere of radius 2r is
Question: The volume of a sphere of radius 2ris (a) $\frac{32 \pi r^{3}}{3}$ (b) $\frac{16 \pi r^{3}}{3}$ (c) $\frac{8 \pi r^{3}}{3}$ (d) $\frac{64 \pi r^{3}}{3}$ Solution: (a) $\frac{32 \pi \mathrm{r}^{3}}{3}$ Volume of the sphere of radius $2 r=\frac{4}{3} \pi(2 r)^{3}=\frac{32}{3} \pi r^{3}$...
Read More →A conical tent is to accommodate 11 persons such that each person occupies 4 m2 of space on the ground.
Question: A conical tent is to accommodate 11 persons such that each person occupies 4 m2of space on the ground. They have 220 m3of air to breathe. The height of the cone is(a) 14 m(b) 15 m(c) 16 m(d) 20 m Solution: (b) 15 mSuppose that the height of the cone ishm. Area of the ground $=11 \times 4=44 \mathrm{~m}^{2}$ $\therefore \pi r^{2}=44$ $\Rightarrow r^{2}=\frac{44 \times 7}{22}=14$ Also, $\frac{1}{3} \pi r^{2} h=220$ $\Rightarrow \frac{1}{3} \times \frac{22}{7} \times 14 h=220$ $\Rightarro...
Read More →The maximum value of the term independent of
Question: The maximum value of the term independent of ' $\mathrm{t}$ ' in the expansion of $\left(t x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{t}\right) \quad$ where $x \in(0,1)$ is:(1) $\frac{10 !}{\sqrt{3}(5 !)^{2}}$(2) $\frac{2.10 !}{3(5 !)^{2}}$(3) $\frac{10 !}{3(5 !)^{2}}$(4) $\frac{2.10 !}{3 \sqrt{3}(5 !)^{2}}$Correct Option: , 4 Solution: $T_{r+1}={ }^{10} C_{r}\left(t x^{1 / 5}\right)^{10-r}\left[\frac{(1-x)^{1 / 10}}{t}\right]^{r}$ $={ }^{10} C_{r} t^{(10-2 r)} \times x^{\frac{10-r}{5...
Read More →Arrange the following metal complex/ compounds in the increasing order of spin only magnetic moment.
Question: Arrange the following metal complex/ compounds in the increasing order of spin only magnetic moment. Presume all the three, high spin system. (Atomic numbers $\mathrm{Ce}=58, \mathrm{Gd}=64$ and $\mathrm{Eu}=63 .)$ (a) $\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{Ce}\left(\mathrm{NO}_{3}\right)_{6}\right]$ (b) $\mathrm{Gd}\left(\mathrm{NO}_{3}\right)_{3}$ and (c) $\mathrm{Eu}\left(\mathrm{NO}_{3}\right)_{3}$$(b)(a)(c)$(c) $(a)(b)$$(a)(b)(c)$$(a)(c)(b)$Correct Option: , 4 Solution: (a...
Read More →A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to make n solid cones of height 1 cm and base radius 1 mm.
Question: A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to makensolid cones of height 1 cm and base radius 1 mm. The value ofnis(a) 450(b) 1350(c) 4500(d) 13500 Solution: (d) 13500Radius of the cylinder = 3 cmHeight of the cylinder = 5 cmRadius of the cone = 1 mm = 0.1 cmHeight of the cone = 1 cm $\therefore$ Number of cones, $n=\frac{\text { volume of the cylinder }}{\text { volume of } 1 \text { cone }}$ $=\frac{\pi \times 3^{2} \times 5}{\frac{1}{3} \pi \times 0.1^{2...
Read More →If the remainder when x is divided by 4 is 3 ,
Question: If the remainder when $x$ is divided by 4 is 3 , then the remainder when $(2020+x)^{2022}$ is divided by 8 is Solution: Let $x=4 k+3$ $(2020+x)^{2022}$ $=(2020+4 k+3)^{2022}$ $=(4(505)+4 k+3)^{2022}$ $=(4 P+3)^{2022}$ $=(4 P+4-1)^{2022}$ $=(4 A-1)^{2022}$ ${ }^{2022} \mathrm{C}_{0}(4 \mathrm{~A})^{0}(-1)^{2022}+{ }^{2022} \mathrm{C}_{1}(4 \mathrm{~A})^{1}(-1)^{2021}+\ldots \ldots$ $1+8 \lambda$ Reminder is 1...
Read More →If the height and the radius of a cone are doubled, the volume of the cone becomes
Question: If the height and the radius of a cone are doubled, the volume of the cone becomes(a) 3 times(b) 4 times(c) 6 times(d) 8 times Solution: (d) 8 timesLet the new height and radius be 2hand 2r, respectively. New volume of the cone $=\frac{1}{3} \pi(2 r)^{2} \times 2 h$ $=\frac{8}{3} \pi r^{2} h=8 \times$ initial volume of the cone...
Read More →The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state.
Question: The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, $\lambda_{1} / \lambda_{2}$, of the photons emitted in this process is :(1) $20 / 7$(2) $27 / 5$(3) $7 / 5$(4) $9 / 7$Correct Option: 1, Solution: (1) $\frac{1}=\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)$ $=\frac{7 R}{16 \times 9}$ And $\frac{1}{\lambda_{2}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\rig...
Read More →When light of wavelength 248 nm falls on
Question: When light of wavelength $248 \mathrm{~nm}$ falls on a metal of threshold energy $3.0 \mathrm{eV}$, the de-Broglie wavelength of emitted electrons is ____________\AA (Round off to the Nearest Integer). $\left[\right.$ Use: $\sqrt{3}=1.73, \mathrm{~h}=6.63 \times 10^{-34} \mathrm{~J}_{\mathrm{S}}$$\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg} ; \mathrm{c}=3.0 \times 10^{8} \mathrm{~ms}^{-1}$ $\left.\mathrm{leV}=1.6 \times 10^{-19} \mathrm{~J}\right]$ Solution: (9) Energy $=\f...
Read More →For integers n and r, let
Question: For integers $n$ and $r$, let $\left(\begin{array}{l}n \\ r\end{array}\right)= \begin{cases}{ }^{n} C_{r}, \text { if } n \geq r \geq 0 \\ 0, \text { otherwise }\end{cases}$ The maximum value of $\mathrm{k}$ for which the sum $\sum_{i=0}^{k}\left(\begin{array}{c}10 \\ i\end{array}\right)\left(\begin{array}{c}15 \\ k-i\end{array}\right)+\sum_{i=0}^{k+1}\left(\begin{array}{c}12 \\ i\end{array}\right)\left(\begin{array}{c}13 \\ k+1-i\end{array}\right)$ exists, is equal to Note: NTA has dr...
Read More →The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3.
Question: The radii of the bases of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. Then, their volumes are in the ratio(a) 9 : 8(b) 8 : 9(c) 3 : 4(d) 4 : 3 Solution: (a) 9 : 8Suppose that the respective radii of the cylinder and the cone are 3rand 4rand their respective heights are 2hand 3h. $\therefore$ Ratio of the volumes $=\frac{\pi(3 r)^{2} \times 2 h}{\frac{1}{3} \pi(4 r)^{2} \times 3 h}=\frac{3 \times 9 \times 2}{16 \times 3}=\frac{9}{8}$...
Read More →A right circular cylinder and a right circular cone have the same radius and the same volume
Question: A right circular cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is(a) 3 : 5(b) 2 : 5(c) 3 : 1(d) 1 : 3 Solution: (d) 1 : 3It is given that the right circular cylinder and the right circular cone have the same radius and volume.Suppose that the radii of their bases are equal torand the respective heights of the cylinder and the cone arehandH.As the volumes of the cylinder and the cone are the same,...
Read More →The ratio of the volumes of a right circular cylinder and a right circular cone of the same base and the same height will be
Question: The ratio of the volumes of a right circular cylinder and a right circular cone of the same base and the same height will be(a) 1 : 3(b) 3 : 1(c) 4 : 3(d) 3 : 4 Solution: (b) 3 : 1It is given that the right circular cylinder and the right circular cone have the same base and height.Suppose that the respective radii of bases and heights are equal torandh. Then ratio of their volumes $=\frac{\pi r^{2} h}{\frac{1}{3} \pi r^{2} h}=\frac{3}{1}$...
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