A cylindrical tub of radius 12 cm contains water to a depth of 20 cm.
Question: A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball? Solution: Suppose that the radius of the ball isrcm.Radius of the cylindrical tub =12 cmDepth of the tub= 20 cmNow, volume of the ball = volume of water raised in the cylinder $\Rightarrow \frac{4}{3} \pi r^{3}=\pi \times 12^{2} \times 6.75$ $\Rightarrow r^{3}=\frac{144 \times 6.75 \times 3}...
Read More →The surface areas of two spheres are in the ratio 1 : 4.
Question: The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes. Solution: Suppose that the radii of the spheres arerandR. We have: $\frac{4 \pi r^{2}}{4 \pi \mathrm{R}^{2}}=\frac{1}{4}$ $\Rightarrow \frac{r}{R}=\sqrt{\frac{1}{4}}=\frac{1}{2}$ Now, ratio of the volumes $=\frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R^{3}}=\left(\frac{r}{R}\right)^{3}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$ The ratio of the volumes of the spheres is 1 : 8....
Read More →Given figure shows few data points in a photo electric effect experiment for a certain metal.
Question: Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is : (Plancks constant $h=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ ) (1) $2.27 \mathrm{eV}$(2) $2.59 \mathrm{eV}$(3) $1.93 \mathrm{eV}$(4) $2.10 \mathrm{eV}$Correct Option: 1 Solution: (1) Graph of $V_{s}$ and $f$ given at $B(5.5,0)$ Minimum energy for ejection of electron $=$ Work function $(\phi)$. $\phi=h V$ joule or $\phi=\...
Read More →The radii of two spheres are in the ratio 1 : 2.
Question: The radii of two spheres are in the ratio 1 : 2. Find the ratio of their surface areas. Solution: Suppose that the radii arerand 2r. Now, ratio of the surface areas $=\frac{4 \pi r^{2}}{4 \pi(2 r)^{2}}=\frac{r^{2}}{4 r^{2}}=\frac{1}{4}$ = 1:4 The ratio of their surface areas is 1 : 4....
Read More →A spherical ball of radius 3 cm is melted and recast into three spherical balls.
Question: A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball. Solution: Radius of the original spherical ball = 3 cmSuppose that the radius of third ball isrcm.Then volume of the original spherical ball = volume of the three spherical balls $\Rightarrow \frac{4}{3} \pi \times 3^{3}=\frac{4}{3} \pi \times 1.5^{3}+\frac{4}{3} \pi \times 2^{3}+\frac{4}{3} \pi \times r^{3}$ $\Rightarr...
Read More →A spherical cannonball 28 cm in diameter is melted and cast into a right circular cone would,
Question: A spherical cannonball 28 cm in diameter is melted and cast into a right circular cone would, whose base is 35 cm in diameter. Find the height of the cone. Solution: Radius of the spherical cannonball,R= 14 cm Radius of the base of the cone,r= 17.5 cmLethcm be the height of the cone.Now, volume of the sphere = volume of the cone $\Rightarrow \frac{4}{3} \pi R^{3}=\frac{1}{3} \pi r^{2} h$ $\Rightarrow 4 \times 14^{3}=(17.5)^{2} \times h$ $\Rightarrow h=\frac{4 \times 14 \times 14 \times...
Read More →A sphere of diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm.
Question: A sphere of diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm. Find the diameter of the base of the cone. Solution: Radius of the sphere,r= 7.8 cmHeight of the cone, h= 31.2 cmSupposethatRcm is the radius of the cone.Now, volume of the sphere = volume of the cone $\Rightarrow \frac{4}{3} \pi r^{3}=\frac{1}{3} \pi R^{2} h$ $\Rightarrow 4 \times(7.8)^{3}=R^{2} \times 31.2$ $\Rightarrow R^{2}=\frac{4 \times 7.8 \times 7.8 \times 7.8}{31.2}=60.84$ $\Rightarro...
Read More →Two sources of light emit X-rays of wavelength
Question: Two sources of light emit X-rays of wavelength $1 \mathrm{~nm}$ and visible light of wavelength $500 \mathrm{~nm}$, respectively. Both the sources emit light of the same power $200 \mathrm{~W}$. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is :(1) $\frac{1}{500}$(2) 250(3) $\frac{1}{250}$(4) 500Correct Option: 1 Solution: (1) Given, Wavelength of X-rays, $\lambda_{1}=1 \mathrm{~nm}=1 \times 10^{-9} \...
Read More →The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross section.
Question: The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter. Solution: Radius of the copper sphere,R= 9 cmLength of the wire,h= 108 m = 10800 cmVolume of the sphere = volume of the wireSuppose thatrcm is the radius of the wire. Then $\frac{4}{3} \pi R^{3}=\pi r^{2} h$ $\Rightarrow \frac{4}{3} \times 9^{3}=r^{2} \times 10800$ $\Rightarrow r^{2}=\frac{4 \times 729}{3 \times 10800}=\frac{...
Read More →The diameter of a sphere is 6 cm.
Question: The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire Solution: Let the length of the wire behcm.Radius of the wire,r= 1 mm = 0.1 cmRadius of the sphere,R= 3 cmNow, volume of the sphere = volume of the cylindrical wire $\Rightarrow \frac{4}{3} \pi R^{3}=\pi r^{2} h$ $\Rightarrow 4 \times 3^{2}=(0.1)^{2} \times h$ $\Rightarrow h=\frac{4 \times 9}{0.1 \times 0.1}=3600 \mathrm{~cm}=36 \mathrm{~m}$ Length of the wire = 36 m...
Read More →How many spheres 12 cm in diameter can be made from a metallic
Question: How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and height 90 cm? Solution: Diameter of each sphere,d=12 cm Radius of each sphere,r=6 cm Volume of each sphere $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi(6)^{3} \mathrm{~cm}^{3}$ Diameter of base of the cylinder, D=8 cmRadius of base of cylinder, R=4 cmHeight of the cylinder,h= 90 cm Volume of the cylinder $=\pi R^{2} h=\pi(4)^{2} \times 90 \mathrm{~cm}^{3}$ Number of spheres $=\frac{\text { volume of...
Read More →If a semiconductor photodiode can detect a photon with a maximum wavelength
Question: If a semiconductor photodiode can detect a photon with a maximum wavelength of $400 \mathrm{~nm}$, then its band gap energy is : Planck's constant, $h=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}$. Speed of light, $\quad c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$(1) $1.1 \mathrm{eV}$(2) $2.0 \mathrm{eV}$(3) $1.5 \mathrm{eV}$(4) $3.1 \mathrm{eV}$Correct Option: , 4 Solution: (4) Given, Wavelength of photon, $\lambda=400 \mathrm{~nm}$ A photodiode can detect a wavelength corresponding ...
Read More →When the wavelength of radiation falling on a metal is changed from
Question: When the wavelength of radiation falling on a metal is changed from $500 \mathrm{~nm}$ to $200 \mathrm{~nm}$, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to :(1) $0.81 \mathrm{eV}$(2) $1.02 \mathrm{eV}$(3) $0.52 \mathrm{eV}$(4) $0.61 \mathrm{eV}$Correct Option: , 2 Solution: (2) Using equation, $=\frac{h c}{\lambda}-\phi$ $K E_{\max }=\frac{h c}{\lambda}-\phi=\frac{h c}{500}-\phi$ ....(1) Again, $3 K E_{\max }=\fr...
Read More →The correct match between Item
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: Solution: According to Bohr's atomic theory : (A) Kinetic energy of electron is $\propto \frac{Z^{2}}{n^{2}}$ (B) The product of velocity (v) of electron and principal quantum number (n). 'vn' $\propto Z^{2...
Read More →A metallic sphere of radius 10.5 cm is melted and then recast into smaller 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 smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained? Solution: Radius of the metallic sphere,r= 10.5 cm Volume of the sphere $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi(10.5)^{3} \mathrm{~cm}^{3}$ Radius of each smaller cone,r2= 3.05 cmHeight of each smaller cone = 3 cm Volume of each smaller cone $=\frac{1}{3} \pi r_{2}{ }^{2} h=\frac{1}{3} \pi(3.05)^{2} \times 3 \mathrm{~cm}^{3}$ Number of co...
Read More →A particle is moving 5 times as fast as an electron.
Question: A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is $1.878 \times 10^{-4}$. The mass of the particle is close to :(1) $4.8 \times 10^{-27} \mathrm{~kg}$(2) $9.1 \times 10^{-31} \mathrm{~kg}$(3) $1.2 \times 10^{-28} \mathrm{~kg}$(4) $9.7 \times 10^{-28} \mathrm{~kg}$Correct Option: , 4 Solution: (4) de Broglie wavelength $\lambda=\frac{h}{m v} \Rightarrow m=\frac{h}{\lambda v}$ Clearly, $m \propto \frac{1...
Read More →A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls,
Question: A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained. Solution: Radius of the solid sphere = 3 cm Volume of the solid sphere $=\frac{4}{3} \pi r^{3}$ $=\frac{4}{3} \times \frac{22}{7} \times 3^{3} \mathrm{~cm}^{3}$ Diameter of the spherical ball = 0.6 cmRadius of the spherical ball = 0.3 cm Volume of the spherical ball $=\frac{4}{3} \times \frac{22}{7} \times(0.3)^{3} \mathrm{~cm}^{3}$ ...
Read More →How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm?
Question: How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm? Solution: Radius of the sphere = 8 cm Volume of the sphere $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7} \times 8^{3}=2145.52 \mathrm{~cm}^{3}$ Radius of one lead ball = 1 cm Volume of one lead ball $=\frac{4}{3} \times \frac{22}{7} \times 1^{3}=4.19 \mathrm{~cm}^{3}$ $\therefore$ Number of lead balls = $\frac{\text { volume of the sphere }}{\text { volume of one lead ball }}=\frac{2145.52}{4....
Read More →In a hydrogen atom the electron makes a transition from
Question: In a hydrogen atom the electron makes a transition from $(n+1)^{\text {th }}$ level to the $n^{\text {th }}$ level. If $n \gg 1$, the frequency of radiation emitted is proportional to :(1) $\frac{1}{n}$(2) $\frac{1}{n^{3}}$(3) $\frac{1}{n^{2}}$(4) $\frac{1}{n^{4}}$Correct Option: , 2 Solution: (2) Total energy of electron in $n^{\text {th }}$ orbit of hydrogen atom $E_{n}=-\frac{R h c}{n^{2}}$ Total energy of electron in $(n+1)^{\text {th }}$ level of hydrogen atom $E_{n+1}=-\frac{R h ...
Read More →How many lead shots, each 3 mm in diameter, can be made from a cuboid with dimensions
Question: How many lead shots, each 3 mm in diameter, can be made from a cuboid with dimensions (12 cm 11 cm 9 cm)? Solution: Here,l= 12 cm,b= 11 cm andh= 9 cm Volume of the cuboid $=l \times b \times h$ $=12 \times 11 \times 9$ $=1188 \mathrm{~cm}^{3}$ Radius of one lead shot $=3 \mathrm{~mm}=\frac{0.3}{2} \mathrm{~cm}$ Volume of one lead shot $=\frac{4}{3} \times \frac{22}{7} \times\left(\frac{0.3}{2}\right)^{3}$ $=\frac{11 \times 9}{7000}$ $=0.014 \mathrm{~cm}^{3}$ $\therefore$ Number of lead...
Read More →The surface area of a sphere is (576π) cm2.
Question: The surface area of a sphere is (576) cm2. Find its volume. Solution: Surface area of the sphere $=(576 \pi) \mathrm{cm}^{2}$ Suppose thatrcm is the radius of the sphere. Then $4 \pi r^{2}=576 \pi$ $\Rightarrow r^{2}=\frac{576}{4}=144$ $\Rightarrow r=12 \mathrm{~cm}$ $\therefore$ Volume of the sphere $=\frac{4}{3} \times \pi \times 12 \times 12 \times 12 \mathrm{~cm}^{3}$ $=2304 \pi \mathrm{cm}^{3}$...
Read More →The final major product of the following reaction is :
Question: The final major product of the following reaction is : Correct Option: Solution: A proton and a $\mathrm{Li}^{3+}$ 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 ____________ $x \times 10^{-1}$. The value of $x$ is [Rounded off to the nearest integer] [Mass of $\mathrm{Li}^{3+}=8....
Read More →Solve this
Question: Note Take $\pi=\frac{22}{7}$, unless stated otherwise. Find the volume of a sphere whose surface area is 154 cm2. Solution: Let the radius of the sphere bercm. Surface area of the sphere =154 cm2 $\therefore 4 \pi r^{2}=154$ $\Rightarrow 4 \times \frac{22}{7} \times r^{2}=154$ $\Rightarrow r=\sqrt{\frac{154 \times 7}{4 \times 22}}=\sqrt{12.25}$ $\Rightarrow r=3.5 \mathrm{~cm}$ $\therefore$ Volume of the sphere $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7} \times(3.5)^{3} \app...
Read More →The area bounded by the lines is
Question: The area bounded by the lines $y=|| x-1|-2|$ is Note: NTA has dropped this question in the final official answer key. Solution: $\mathrm{A}=\int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}}(\sin \mathrm{x}-\cos \mathrm{x}) \mathrm{dx}=[-\cos \mathrm{x}-\sin \mathrm{x}]_{\pi / 4}^{5 \pi / 4}$ $=-\left[\left(\cos \frac{5 \pi}{4}+\sin \frac{\pi}{4}\right)-\left(\cos \frac{\pi}{4}+\sin \frac{\pi}{4}\right)\right]$ $=-\left[\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)-\left(\frac{1}{\sqrt{2}}+\...
Read More →When radiation of wavelength
Question: When radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is $V$. When the same surface is illuminated with radiation of wavelength $3 \lambda$, the stopping potential is $\frac{V}{4} .$ If the theshold wavelength for the metallic surface is $n \lambda$ then value of $n$ will be_______ Solution: $(9)$ When radiation of wavelength $A, \lambda_{A}$ is used to illuminate, stopping potential $V_{A}=V$ $\frac{h c}{\lambda}=\phi+e V$ ...(1) When ...
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