The plots of radial distribution functions for various orbitals of hydrogen atom against
Question: The plots of radial distribution functions for various orbitals of hydrogen atom against ' $r$ ' are given below: The correct plot for $3 \mathrm{~s}$ orbital is:DBACCorrect Option: 1 Solution: $3 s$ orbital Number of radial nodes $=\mathrm{n}-\ell-1$ For $3 \mathrm{~s}$ orbital $n=3 \quad \ell=0$ Number of radial nodes $=3-0-1=2$ It is correctly represented in graph of option $\mathrm{D}$...
Read More →The coefficient of
Question: The coefficient of $x^{4}$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{6}$ in powers of $x$, is__________. Solution: Coefficient of $x^{4}$ in $\left(\frac{1-x^{4}}{1-x}\right)^{6}=$ coefficient of $x^{4}$ in $\left(1-6 x^{4}\right)(1-x)^{-6}$ $=$ coefficient of $x^{4}$ in $\left(1-6 x^{4}\right)\left[1+{ }^{6} C_{1} x+{ }^{7} C_{2} x^{2}+\ldots . .\right]$ $={ }^{9} C_{4}-6 \cdot 1=126-6=120$...
Read More →A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm.
Question: A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. The radius of the base of the cone is(a) 6.3 cm(b) 2.1 cm(c) 6 cm(d) 4 cm Solution: (a) 6.3 cmLetrcm be the radius of base of the cone.Volume of the sphere = volume of the cone $\Rightarrow \frac{4}{3} \pi \times(6.3)^{3}=\frac{1}{3} \pi \times r^{2} \times 25.2$ $\Rightarrow r^{2}=\frac{4 \times 6.3 \times 6.3 \times 6.3}{25.2}=39.69$ $\Rightarrow r=\sqrt{39.69}=6.3 \mathrm{~cm}$...
Read More →The natural number m,
Question: The natural number $m$, for which the coefficient of $x$ in the binomial expansion of $\left(x^{m}+\frac{1}{x^{2}}\right)^{22}$ is 1540 , is__________. Solution: $T_{r+1}={ }^{22} C_{r} \cdot\left(x^{m}\right)^{22-r} \cdot\left(\frac{1}{x^{2}}\right)^{r}$ $T_{r+1}={ }^{22} C_{r} \cdot x^{22 m-m r-2 r}$ $\because 22 m-m r-2 r=1$ $\Rightarrow r=\frac{22 m-1}{m+2} \Rightarrow r=22-\frac{3 \cdot 3 \cdot 5}{m+2}$ So, possible value of $m=1,3,7,13,43$ But ${ }^{22} C_{r}=1540$ $\therefore$ O...
Read More →In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell,
Question: In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $\mathrm{N}$-shell to the $\mathrm{L}$-shell, the wavelength of emitted radiation will be:(1) $\frac{27}{20} \lambda$(2) $\frac{16}{25} \lambda$(3) $\frac{25}{16} \lambda$(4) $\frac{20}{27} \lambda$Correct Option: , 4 Solution: (4) When electron jumps from $M \rightarrow L$ shell $\frac{1}{\lambda}=K\left(\frac{1}{2^{2}}-\frac{1}{3...
Read More →The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm.
Question: The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is(a) 12 m(b) 18 m(c) 36 m(d) 66 m Solution: (c) 36 mLethm be the length of the wire.Volume of the sphere = volume of the wire $\Rightarrow \frac{4}{3} \pi \times 3^{3}=\pi \times(0.1)^{2} \times h$ $\Rightarrow h=\frac{4 \times 9}{0.001}$ $=3600 \mathrm{~cm}$ $=36 \mathrm{~m}$...
Read More →How many lead shots, each 0.3 cm in diameter, can be made from a cuboid of dimensions 9 cm × 11 cm× 12 cm?
Question: How many lead shots, each 0.3 cm in diameter, can be made from a cuboid of dimensions 9 cm11 cm 12 cm?(a) 7200(b) 8400(c) 72000(d) 84000 Solution: (d) 84000 Number of lead shots $=\frac{\text { volume of cuboid }}{\text { volume of } 1 \text { lead shot }}$ $=\frac{9 \times 11 \times 12}{\frac{4}{3} \times \frac{22}{7} \times 0.15 \times 0.15 \times 0.15}$ $=\frac{9 \times 11 \times 3 \times 3 \times 7}{22 \times 0.15 \times 0.15 \times 0.15}$ $=84000$...
Read More →If for some positive integer n,
Question: If for some positive integer $n$, the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+5}$ are in the ratio $5: 10: 14$, then the largest coefficient in this expansion is :(1) 462(2) 330(3) 792(4) 252Correct Option: 1 Solution: Consider the three consecutive coefficients of $(1+x)^{n+5}$ be ${ }^{n+5} C_{r},{ }^{n+5} C_{r+1},{ }^{n+5} C_{r+2}$ \because \frac{{ }^{n+5} C_{r}}{{ }^{n+5} C_{r+1}}=\frac{1}{2} (Given) $\Rightarrow \frac{r+1}{n+5-r}=\frac{1}{2} ...
Read More →A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm.
Question: A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Then number of such cones will be(a) 21(b) 63(c) 126(d) 130 Solution: (c) 126 Number of cones $=\frac{\text { volume of the sphere }}{\text { volume of } 1 \text { cone }}$ $=\frac{\frac{4}{3} \pi \times 10.5 \times 10.5 \times 10.5}{\frac{1}{3} \pi \times 3.5 \times 3.5 \times 3}$ $=4 \times 3 \times 3 \times 3.5=126$...
Read More →According to Bohr's atomic theory :
Question: According to Bohr's atomic theory : (A) Kinetic energy of electron is $\propto \frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}$ (B) The product of velocity (v) of electron and principal quantum number (n). 'vn' $\propto \mathrm{Z}^{2}$. (C) Frequency of revolution of electron in an orbit is $\propto \frac{Z^{3}}{n^{3}}$. (D) Coulombic force of attraction on the electron is $\propto \frac{Z^{3}}{n^{4}}$. Choose the most appropriate answer from the options given below:(C) only(A) and (D) only(A) o...
Read More →If the deBroglie wavelength of an electron is equal
Question: If the deBroglie wavelength of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6 \times 10^{14} \mathrm{~Hz}$, then the speed of electron is equal to : (Speed of light $=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ) Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ )(1) $1.1 \times 10^{6} \mathrm{~m} / \mathrm{s}$(2) $1.7 \times 10^{6} \mathrm{~m} / \mathrm{s}$(3) $1.8 \times 10^{6} \ma...
Read More →A solid lead ball of radius 6 cm is melted and then drawn into a wire of diameter 0.2 cm.
Question: A solid lead ball of radius 6 cm is melted and then drawn into a wire of diameter 0.2 cm. The length of wire is(a) 272 m(b) 288 m(c) 292 m(d) 296 m Solution: (b) 288 mLethm be the length of wire.Volume of the lead ball = volume of the wire $\Rightarrow \frac{4}{3} \pi \times 6^{3}=\pi \times(0.1)^{2} h$ $\Rightarrow h=\frac{4 \times 216}{3 \times 0.01}=28800 \mathrm{~cm}=288 \mathrm{~m}$...
Read More →Let
Question: Let $\left(2 x^{2}+3 x+4\right)^{10}=\sum_{r=0}^{20} a_{r} x^{r}$. Then $\frac{a_{7}}{a_{13}}$ is equal to___________. Solution: The given expression is $\left(2 x^{2}+3 x+4\right)^{10}=\sum_{r=0}^{20} a_{r} x^{r}$ General term $=\frac{10 !}{r_{1} ! r_{2} ! r_{3} !}\left(2 x^{2}\right)^{\eta}(3 x)^{r_{2}}(4)^{r_{3}}$ Since, $a_{7}=$ Coeff. of $x^{7}$ $2 r_{1}+r_{2}=7$ and $r_{1}+r_{2}+r_{3}=10$ Possibilities are $a_{7}=\frac{10 ! 3^{7} 4^{3}}{7 ! 3 !}+\frac{10 !(2)(3)^{5}(4)^{4}}{5 ! 4...
Read More →A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength
Question: A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength $980 \AA$. The radius of the atom in the excited state, in terms of Bohr radius $\mathrm{a}_{0}$, will be: $(h c=12500 \mathrm{eV}-A)$(1) $25 \mathrm{a}_{0}$(2) $9 \mathrm{a}_{0}$(3) $16 \mathrm{a}_{0}$(4) $4 \mathrm{a}_{0}$Correct Option: , 3 Solution: (3) Energy of photon $=\frac{\text { hc }}{\lambda}=\frac{12500}{980}=12.75 \mathrm{eV}$ Energy of electron in $\mathrm{n}^{\text {th }}$ orbi...
Read More →A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere.
Question: A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is(a) 4.2 cm(b) 2.1 cm(c) 2.4 cm(d) 1.6 cm Solution: (b) 2.1 cmLetrcm be the radius of the sphere.Volume of the cone = volume of the sphere $\Rightarrow \frac{1}{3} \pi \times 2.1 \times 2.1 \times 8.4=\frac{4}{3} \pi r^{3}$ $\Rightarrow r^{3}=2.1 \times 2.1 \times 2.1$ $\Rightarrow r=2.1 \mathrm{~cm}$...
Read More →Solve the following
Question: A proton and $\mathrm{aLi}^{3+}$ nucleus are accelerated by the same potential. If $\lambda_{\mathrm{Li}}$ and $\lambda_{\mathrm{p}}$ denote the de Broglie wavelengths of $\mathrm{Li}^{3+}$ and proton respectively, then the value of $\frac{\lambda_{\mathrm{Li}}}{\lambda_{\mathrm{p}}}$ is $\mathrm{x} \times 10^{-1}$. The value of $x$ is _________ [Rounded off to the nearest integer] [Mass of $\mathrm{Li}^{3+}=8.3$ mass of proton] Solution: (2) De Brogir Davelength $\lambda=\frac{\mathrm...
Read More →A solid metal ball of radius 8 cm is melted and cast into smaller balls, each of radius 2 cm.
Question: A solid metal ball of radius 8 cm is melted and cast into smaller balls, each of radius 2 cm. The number of such balls is(a) 8(b) 16(c) 32(d) 64 Solution: (d) 64 Number of balls $=\frac{\text { volume of solid metal ball }}{\text { volumeof } 1 \text { small ball }}$ $=\frac{\frac{4}{3} \pi \times 8^{3}}{\frac{4}{3} \pi \times 2^{3}}=\frac{512}{8}=64$...
Read More →Given below are two statements:
Question: Given below are two statements: Statement I : Bohr's theory accounts for the stability and line spectrum of $\mathrm{Li}^{+}$ion. Statement II : Bohr's theory was unable to explain the splitting of spectral lines in the presence of a magnetic field. In the light of the above statements, choose the most appropriate answer from the options given below:Both statement I and statement II are true.Statement $\mathrm{I}$ is false but statement $\mathrm{II}$ is true.Both statement I and statem...
Read More →If the ratio of the volumes of two spheres is 1 : 8,
Question: If the ratio of the volumes of two spheres is 1 : 8, then the ratio of their surface area is(a) 1 : 2(b) 1 : 4(c) 1 : 8(d) 1 : 16 Solution: (b) 1 : 4SupposethatrandRare the radii of the spheres.Then we have: $\frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R^{3}}=\frac{1}{8}$ $\Rightarrow\left(\frac{r}{R}\right)^{3}=\frac{1}{8}$ $\Rightarrow \frac{r}{R}=\frac{1}{2}$ $\therefore$ Ratio of surface area of spheres $=\frac{4 \pi r^{2}}{4 \pi R^{2}}=\left(\frac{r}{R}\right)^{2}=\left(\frac{1}{...
Read More →Surface of certain metal is first illuminated with light of wavelength
Question: Surface of certain metal is first illuminated with light of wavelength $\lambda_{1}=350 \mathrm{~nm}$ and then, by light of wavelength $\lambda_{2}=540 \mathrm{~nm}$. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of (2) The work function of the metal (in $\mathrm{eV}$ ) is close to: (Energy of photon $\left.=\frac{1240}{\lambda(\text { in } \mathrm{nm})} \mathrm{eV}\right)$(1) $1.8$(2) $2.5$(3) $5.6$(4) $1.4$Correct Option: 1 Solution: (1...
Read More →The value of
Question: The value of $\sum_{r=0}^{20}{ }^{50-r} C_{6}$ is equal to:(1) ${ }^{51} C_{7}-{ }^{30} C_{7}$(2) ${ }^{50} C_{7}-{ }^{30} C_{7}$(3) ${ }^{50} C_{6}-{ }^{30} C_{6}$(4) ${ }^{51} C_{7}+{ }^{30} C_{7}$Correct Option: 1 Solution: The given series, $\sum_{r=0}^{20}{ }^{50-r} C_{6}$ $={ }^{50} C_{6}+{ }^{49} C_{6}+{ }^{48} C_{6}+{ }^{47} C_{6}+\ldots+{ }^{32} C_{6}+{ }^{31} C_{6}+{ }^{30} C_{6}$ $=\left({ }^{30} C_{7}+{ }^{30} C_{6}\right)+{ }^{31} C_{6}+{ }^{32} C_{6}+\ldots .+{ }^{49} C_{...
Read More →The volume of a sphere is 38808 cm3.
Question: The volume of a sphere is 38808 cm3. Its curved surface area is(a) 5544 cm2(b) 8316 cm2(c) 4158 cm2(d) 1386 cm2 Solution: (a) $5544 \mathrm{~cm}^{2}$ Letrcm be the radius of the sphere.Then we have: $\frac{4}{3} \pi r^{3}=38808$ $\Rightarrow r^{3}=\frac{38808 \times 3 \times 7}{4 \times 22}=9261$ $\Rightarrow r=21 \mathrm{~cm}$ $\therefore$ Curved surface area $=4 \pi r^{2}=4 \times \frac{22}{7} \times 21 \times 21$ $=5544 \mathrm{~cm}^{3}$...
Read More →A certain orbital has no angular nodes and two radial nodes.
Question: A certain orbital has no angular nodes and two radial nodes. The orbital is:$2 \mathrm{~s}$$3 \mathrm{~s}$$3 \mathrm{p}$$2 \mathrm{p}$Correct Option: , 2 Solution: $1=0 \Rightarrow$ 's' orbital $\mathrm{n}-l-1=2$ $\operatorname{lnh}-1=2$ $\mathrm{n}=3$...
Read More →If the term independent of x in the expansion
Question: If the term independent of $x$ in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ is $k$, then $18 k$ is equal to :(1) 5(2) 9(3) 7(4) 11Correct Option: , 3 Solution: General term $=T_{r+1}={ }^{9} C_{r}\left(\frac{3 x^{2}}{2}\right)^{9-r}\left(-\frac{1}{3 x}\right)^{r}$ $={ }^{9} C_{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^{r} x^{18-3 r}$ The term is independent of $x$, then $18-3 r=0 \Rightarrow r=6$ $\therefore T_{7}={ }^{9} C_{6}\left(\frac{3}{2}...
Read More →In the ground state of atomic
Question: In the ground state of atomic $\mathrm{Fe}(\mathrm{Z}=26)$, the spin-only magnetic moment is ____________ $\times 10^{-1} \mathrm{BM}$. (Round off to the Nearest Integer). [Given : $\sqrt{3}=1.73, \sqrt{2}=1.41]$ Solution: (49) Number of unpaired $\mathrm{e}^{-}=4$ $\mu=\sqrt{4(4+2)} \mathrm{B} \cdot \mathrm{M}$ $\mu=\sqrt{24} \mathrm{~B} \cdot \mathrm{M}$ $\mu=4.89 \mathrm{~B} \cdot \mathrm{M}$ $\mu=48.9 \times 10^{-1} \mathrm{~B} \cdot \mathrm{M}$ Nearest integer value will be 49 ....
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