2.2 dm3 of lead is to be drawn into a cylindrical wire 0.50 cm in diameter.
Question: 2.2 dm3of lead is to be drawn into a cylindrical wire 0.50 cm in diameter. The length of the wire is(a) 110 m(b) 112 m(c) 98 m(d) 124 m Solution: (b) 112 m Volume of the lead = volume of the cylindrical wireSupposethathm is the length of the wire. As $2.2 \mathrm{dm}^{3}$ of lead is to be drawn into a cylindrical wire of diameter $0.50 \mathrm{~cm}$, we have: $\pi(0.25)^{2} \times h=2.2 \times 10 \times 10 \times 10$ $\Rightarrow h=\frac{2200 \times 7}{22 \times 0.0625}$ $=\frac{700}{0...
Read More →If the fourth term in the expansion of
Question: If the fourth term in the expansion of $\left(\mathrm{x}+\mathrm{x}^{\log _{2} \mathrm{x}}\right)^{7}$ is 4480 , then the value of $x$ where $x \in N$ is equal to:(1) 2(2) 4(3) 3(4) 1Correct Option: 1 Solution: ${ }^{7} \mathrm{C}_{3} \mathrm{x}^{4} \mathrm{x}\left({ }^{3 \log _{2}^{2}}\right)=4480$ $\Rightarrow x\left(4+3 \log _{2}^{2}\right)=2^{7}$ $\Rightarrow \quad(4+3 \mathrm{t}) \mathrm{t}=7 ; \mathrm{t}=\log _{2}^{\mathrm{x}}$ $\Rightarrow \mathrm{t}=1, \frac{-7}{3} \Rightarrow ...
Read More →The correct match between Item - I (starting material) and Item - II (reagent)
Question: The correct match between Item - I (starting material) and Item - II (reagent) for the preparation of benzaldehyde is : (I) - (Q), (II) - (R) and (III) - (P)(I) - (P), (II) - (Q) and (III) - (R)(I) - (R), (II) - (P) and (III) - (Q)(I) - (R), (II) - (Q) and (III) - (P)Correct Option: , 3 Solution:...
Read More →A proton, an electron, and a Helium nucleus, have the same energy.
Question: A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let $\mathrm{r}_{\mathrm{p}}, \mathrm{r}_{\mathrm{e}}$ and $\mathrm{r}_{\mathrm{He}}$ be their respective radii, then, (1) $\mathrm{r}_{\mathrm{e}}\mathrm{r}_{\mathrm{p}}=\mathrm{r}_{\mathrm{He}}$(2) $\mathrm{r}_{\mathrm{e}}\mathrm{r}_{\mathrm{p}}=\mathrm{r}_{\mathrm{He}}$(3) $\mathrm{r}_{\mathrm{e}}\mathrm{r}_{\mathrm{p}}\mathrm{r}...
Read More →The diameter of a roller, 1 m long, is 84 cm. If it takes 500 complete revolutions to level a playground,
Question: The diameter of a roller, 1 m long, is 84 cm. If it takes 500 complete revolutions to level a playground, the area of the playground is(a) 1440 m2(b) 1320 m2(c) 1260 m2(d) 1550 m2 Solution: (b) 1320 m2 Area covered by the roller in 1 revolution $=2 \pi r h$ $=2 \times \frac{22}{7} \times 42 \times 100$ $=44 \times 600$ $=26400 \mathrm{~cm}^{2}$ $\therefore$ Area covered in 50 revolutions $=26400 \times 500 \mathrm{~cm}^{2}$ $=1320 \mathrm{~m}^{2}$...
Read More →Let n be a positive integer. Let
Question: Let $\mathrm{n}$ be a positive integer. Let\ $\mathrm{A}=\sum_{\mathrm{k}=0}^{\mathrm{n}}(-1)^{\mathrm{k}} \mathrm{n}_{\mathrm{C}_{\mathrm{k}}}\left[\left(\frac{1}{2}\right)^{\mathrm{k}}+\left(\frac{3}{4}\right)^{\mathrm{k}}+\left(\frac{7}{8}\right)^{\mathrm{k}}+\left(\frac{15}{16}\right)^{\mathrm{k}}+\left(\frac{31}{32}\right)^{\mathrm{k}}\right]$ If $63 \mathrm{~A}=1-\frac{1}{2^{30}}$, then $\mathrm{n}$ is equal to Solution: $\mathrm{A}=\sum_{k=0}^{n}{ }^{n} \mathrm{C}_{\mathrm{k}}\l...
Read More →The radius of a wire is decreased to one-third. If volume remains the same, the length will become
Question: The radius of a wire is decreased to one-third. If volume remains the same, the length will become(a) 2 times(b) 3 times(c) 6 times(d) 9 times Solution: (d) 9 times Let the new radius be $\frac{1}{3} r$. Suppose that the new height isH.The volume remains the same. i. e., $\pi r^{2} h=\pi \times\left(\frac{1}{3} r\right)^{2} \times H$ $\Rightarrow h=\frac{1}{9} H$ $\Rightarrow H=9 h$ $\therefore$ The new height becomes nine times the original height....
Read More →The final major product of the following reaction is :
Question: The final major product of the following reaction is : Correct Option: , 4 Solution:...
Read More →Let [x] denote greatest integer less than or equal to
Question: Let $[x]$ denote greatest integer less than or equal to $x$. If for $n \in \mathbb{N},\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$, then $\left.\sum_{j=0}^{\left[\frac{3 n}{2}\right]} a_{2 j}+4 \sum_{j=0}^{\left[\frac{3 n-1}{2}\right.}\right] a_{2 j}+1$ is equal to :(1) 2(2) $2^{n-1}$(3) 1(3) 1Correct Option: , 3 Solution: $\left(1-x+x^{3}\right)^{n}=\sum_{j=0}^{3 n} a_{j} x^{j}$ $\left(1-x+x^{3}\right)^{n}=a_{0}+a_{1} x+a_{2} x^{2} \ldots \ldots+a_{3 n} x^{3 n}$ $\sum_{j=0...
Read More →The number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular
Question: The number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is(a) 540(b) 450(c) 380(d) 472 Solution: (b) 450 Number of coins $=\frac{\text { volume of cylinder to be formed }}{\text { volume of a coin }}$ $=\frac{\pi \times 2.25 \times 2.25 \times 10}{\pi \times 0.75 \times 0.75 \times 0.2}$ $=\frac{225 \times 225 \times 5}{75 \times 75}$ $=450$...
Read More →In a photoelectric effect experiment the threshold wavelength of light is 380 nm.
Question: In a photoelectric effect experiment the threshold wavelength of light is $380 \mathrm{~nm}$. If the wavelength of incident light is $260 \mathrm{~nm}$, the maximum kinetic energy of emitted electrons will be: Given $\mathrm{E}($ in $\mathrm{eV})=\frac{1237}{\lambda(\text { in } \mathrm{nm})}$(1) $1.5 \mathrm{eV}$(2) $3.0 \mathrm{eV}$(3) $4.5 \mathrm{eV}$(4) $15.1 \mathrm{eV}$Correct Option: 1 Solution: (1) $\mathrm{KE}_{\max }=\mathrm{E}-\phi_{0}$ (where $\mathrm{E}=$ energy of incide...
Read More →In a cylinder, if the radius is halved and the height is doubled, then the volume will be
Question: In a cylinder, if the radius is halved and the height is doubled, then the volume will be(a) the same(b) doubled(c) halved(d) four times Solution: (c) halved Suppose that the new radius is $\frac{1}{2} r$ and the height is $2 h$. $\therefore$ Volume $=\pi \times\left(\frac{1}{2} r\right)^{2} \times 2 h$ $=\pi \times \frac{r^{2}}{4} \times 2 h$ $=\frac{1}{2} \pi r^{2} h$...
Read More →In the following reaction sequence the major products A and $mathrm{B}$ are :
Question: In the following reaction sequence the major products A and $\mathrm{B}$ are : Correct Option: Solution:...
Read More →A He^{+}ion is in its first excited state. Its ionization energy is:
Question: $\mathrm{AHe}^{+}$ion is in its first excited state. Its ionization energy is:(1) $48.36 \mathrm{eV}$(2) $54.40 \mathrm{eV}$(3) $13.60 \mathrm{eV}$(4) $6.04 \mathrm{eV}$Correct Option: , 3 Solution: (3) $E_{n}=-13.6 \frac{Z^{2}}{n^{2}}$ For $\mathrm{He}^{+}, E_{2}=\frac{-13.6(2)^{2}}{2^{2}}=-13.60 \mathrm{eV}$ Ionization energy $=0-E_{2}=13.60 \mathrm{eV}$...
Read More →If n is the number of irrational terms in the expansion o
Question: If $\mathrm{n}$ is the number of irrational terms in the expansion of $\left(3^{1 / 4}+5^{1 / 8}\right)^{60}$, then $(\mathrm{n}-1)$ is divisible by:(1) 26(2) 30(3) 8(4) 7Correct Option: 1 Solution: For rational terms. $\frac{r}{8}=k ; 0 \leq r \leq 60$ $0 \leq 8 k \leq 60$ $0 \leq k \leq \frac{60}{8}$ $0 \leq k \leq 7.5$ $\mathrm{k}=0,1,2,3,4,5,6,7$ $\frac{60-8 \mathrm{k}}{4}$ is always divisible by 4 for all value of $\mathrm{k}$ Total rational terms $=8$ Total terms $=61$ irrational...
Read More →The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2.
Question: The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. If the total surface area is 616 cm2, then the volume of the cylinder is(a) 1078 cm3(b) 1232 cm3(c) 1848 cm3(d) 924 cm3 Solution: (a) 1078 cm3 We have: $\frac{2 \pi r h}{2 \pi r h+2 \pi r^{2}}=\frac{1}{2}$ $\Rightarrow 4 \pi r h=2 \pi r h+2 \pi r^{2}$ $\Rightarrow 2 \pi r h=2 \pi r^{2}$ $\Rightarrow \frac{r}{h}=\frac{1}{1}$ Also, $2 \pi r h+2 \pi r^{2}=616$ $\Rightarrow 2 \pi r^{...
Read More →Taking the wavelength of first Balmer line in hydrogen spectrum
Question: Taking the wavelength of first Balmer line in hydrogen spectrum $(\mathrm{n}=3$ to $\mathrm{n}=2)$ as $660 \mathrm{~nm}$, the wavelength of the $2^{\text {nd }}$ Balmer line $(n=4$ to $n=2)$ will be;(1) $889.2 \mathrm{~nm}$(2) $488.9 \mathrm{~nm}$(3) $642.7 \mathrm{~nm}$(4) $388.9 \mathrm{~nm}$Correct Option: , 2 Solution: (2) $\frac{1}{\lambda_{1}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 R}{36}$ $\frac{1}{\lambda_{2}}=R\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3...
Read More →Two circular cylinders of equal volume have their heights in the ratio 1 : 2.
Question: Two circular cylinders of equal volume have their heights in the ratio 1 : 2. The ratio of their radii is (a) $1: \sqrt{2}$ (b) $\sqrt{2}: 1$ (c) 1 : 2(d) 1 : 4 Solution: (b) $\sqrt{2}: 1$ Suppose that the heights of two cylinders arehand 2hwhose radii arerandR, respectively.Since the volumes of the cylinders are equal, we have: $\pi r^{2} h=\pi R^{2} \times 2 h$ $\Rightarrow \frac{r^{2}}{R^{2}}=\frac{2}{1}$ $\Rightarrow\left(\frac{r}{R}\right)=\sqrt{\frac{2}{1}}$ $\Rightarrow r: R=\sq...
Read More →Solve the following
Question: $[\mathrm{P}]$ on treatment with $\mathrm{Br}_{2} / \mathrm{FeBr}_{3}$ in $\mathrm{CCl}_{4}$ produced a single isomer $\mathrm{C}_{8} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{Br}$ while heating [ $\mathrm{P}$ ] with sodalime gave toluene. The compound [P] is :Correct Option: 1 Solution:...
Read More →The ratio between the radius of the base and the height of a cylinder is 2 : 3. If its volume is 1617 cm3,
Question: The ratio between the radius of the base and the height of a cylinder is 2 : 3. If its volume is 1617 cm3, then its total surface area is(a) 308 cm2(b) 462 cm2(c) 540 cm2(d) 770 cm2 Solution: (d) 770 cm2We have:r: h= 2 : 3 $\Rightarrow \frac{r}{h}=\frac{2}{3}$ $\Rightarrow h=\frac{3}{2} r$ Now, volume $=1617 \mathrm{~cm}^{3}$ $\Rightarrow \pi r^{2} h=1617$ $\Rightarrow \frac{22}{7} \times r^{2} \times \frac{3}{2} r=1617$ $\Rightarrow r^{3}=\frac{1617 \times 14}{66}=343$ $\Rightarrow r=...
Read More →The major product of the following reaction is :
Question: The major product of the following reaction is : Correct Option: , 3 Solution: It is an example of electrophilic substitution reaction. Position of electrophile is directed by the strong ring activating group $(-\mathrm{OH})$, present in the ring....
Read More →Radiation coming from transitions n=2 to n=1 of hydrogen atoms fall on
Question: Radiation coming from transitions $n=2$ to $n=1$ of hydrogen atoms fall on $\mathrm{He}^{+}$ions in $n=1$ and $n=2$ states. The possible transition of helium ions as they absorb energy from the radiation is :(1) $n=2 \rightarrow n=3$(2) $n=1 \rightarrow n=4$(3) $n=2 \rightarrow n=5$(4) $n=2 \rightarrow n=4$Correct Option: , 4 Solution: (4) Energy released by hydrogen atom for transition $n=2$ to $n=1$ $\therefore \Delta E_{1}=13.6 \times\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=\fra...
Read More →The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3.
Question: The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is(a) 27 : 20(b) 20 : 27(c) 4 : 9(d) 9 : 4 Solution: (b) 20 : 27Suppose that the radii of the cylinders are 2rand 3rand their respective heights are 5hand 3h.. $\therefore$ Ratio of their volumes $=\frac{\pi(2 r)^{2} \times 5 h}{\pi(3 r)^{2} \times 3 h}$ $=\frac{4 \times 5}{9 \times 3}=\frac{20}{27}$...
Read More →If n is the number of irrational terms in the expansion of
Question: If $\mathrm{n}$ is the number of irrational terms in the expansion of $\left(3^{1 / 4}+5^{1 / 8}\right)^{60}$, then $(\mathrm{n}-1)$ is divisible by: (1) 26 (2) 30(3) 8(4) 7Correct Option: 1 Solution: Here $\mathrm{AO}+\mathrm{OD}-1$ or $(\sqrt{2}+1) \mathrm{r}=1$ $\Rightarrow \quad \mathbf{r}=\sqrt{2-1}$ equation of circle $(x-r)^{2}+(y-r)^{2}=r^{2}$ Equation of CE $y-1-m(x-1)$ $m x-y+1-M=0$ It is tangent to circle $\therefore \quad\left|\frac{m r-r+1-m}{\sqrt{m^{2}+1}}\right|=r$ $\le...
Read More →The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3.
Question: The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their curved surface area is(a) 2 : 5(b) 8 : 7(c) 10 : 9(d) 16 : 9 Solution: (c) 10 : 9Suppose that the radii of the cylinders are 2rand 3rand their respective heights are 5hand 3h. Then, ratio of the curved surface areas $=\frac{2 \pi(2 r)(5 h)}{2 \pi(3 r)(3 h)}=10: 9$...
Read More →