In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm.
Question: In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle is (a) 1 cm(b) 2 cm(c) 3 cm(d) 4 cm Solution: In right triangle ABCBy using Pythagoras theorem we haveAC2= AB2+ BC2= 52+ 122= 25 + 144= 169 AC2= 169⇒ AC = 13 cmNow, $\operatorname{Ar}(\triangle \mathrm{ABC})=\operatorname{Ar}(\triangle \mathrm{AOB})+\operatorname{Ar}(\triangle \mathrm{BOC})+\operatorname{Ar}(\triangle \mathrm{AOC})$ $\Rightarrow \frac{1}{2} \times \m...
Read More →A solid piece of iron of dimensions 49 × 33 × 24 cm
Question: A solid piece of iron of dimensions 49 33 24 cm is moulded into a sphere. The radius of the sphere is(a) 21 cm(b) 28 cm(c) 35 cm(d) none of these Solution: The volume of iron piece = 49 33 24 cm3 Let,ris the radius sphere. Clearly, The volume of sphere = volume of iron piece $\frac{4}{3} \pi r^{3}=49 \times 33 \times 24$ $\frac{4}{3} \times \frac{22}{7} \times r^{3}=49 \times 33 \times 24$ $r^{3}=49 \times 3 \times 3 \times 3 \times 7$ $r=7 \times 3$ $r=21 \mathrm{~cm}$ Hence, the corr...
Read More →A solid piece of iron of dimensions 49 × 33 × 24 cm
Question: A solid piece of iron of dimensions 49 33 24 cm is moulded into a sphere. The radius of the sphere is(a) 21 cm(b) 28 cm(c) 35 cm(d) none of these Solution: The volume of iron piece = 49 33 24 cm3 Let,ris the radius sphere. Clearly, The volume of sphere = volume of iron piece $\frac{4}{3} \pi r^{3}=49 \times 33 \times 24$ $\frac{4}{3} \times \frac{22}{7} \times r^{3}=49 \times 33 \times 24$ $r^{3}=49 \times 3 \times 3 \times 3 \times 7$ $r=7 \times 3$ $r=21 \mathrm{~cm}$ Hence, the corr...
Read More →A solid piece of iron of dimensions 49 × 33 × 24 cm
Question: A solid piece of iron of dimensions 49 33 24 cm is moulded into a sphere. The radius of the sphere is(a) 21 cm(b) 28 cm(c) 35 cm(d) none of these Solution: The volume of iron piece = 49 33 24 cm3 Let,ris the radius sphere. Clearly, The volume of sphere = volume of iron piece $\frac{4}{3} \pi r^{3}=49 \times 33 \times 24$ $\frac{4}{3} \times \frac{22}{7} \times r^{3}=49 \times 33 \times 24$ $r^{3}=49 \times 3 \times 3 \times 3 \times 7$ $r=7 \times 3$ $r=21 \mathrm{~cm}$ Hence, the corr...
Read More →In the given figure, O is the centre of the circle AB is a chord and AT is the tangent at A.
Question: In the given figure, O is the centre of the circle AB is a chord and AT is the tangent at A. If AOB = 100∘then BAT is equal to (a) 40∘(b) 50∘(c) 90∘(d) 100∘ Solution: Given: AO and BO are the radius of the circleSince, AO = BO △AOB is an isosceles triangle.Now, in △AOBAOB + OBA + OAB = 180∘ (Angle sum property of triangle)⇒ 100∘ + OAB + OAB = 180∘ (OBA = OAB)⇒ 2OAB = 80∘⇒ OAB = 40∘We know that the radius and tangent are perperpendular at their point of contact∵OAT = 90∘⇒ OAB + BAT = 90...
Read More →A hollow sphere of internal and external diameters 4 cm and 8 cm
Question: A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. The height of the cone is(a) 12 cm(b) 14 cm(c) 15 cm(d) 18 cm Solution: External radius $r_{\mathrm{i}}=\frac{8}{2}=4 \mathrm{~cm}$ Internal radius $r_{2}=\frac{4}{2}=2 \mathrm{~cm}$ The volume of hollow sphere $V=\frac{4}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right)$ $=\frac{4}{3} \pi\left(4^{3}-2^{3}\right)$ Lethbe the height of cone. Clearly, The volume of r...
Read More →A solid metallic spherical ball of diameter 6 cm
Question: A solid metallic spherical ball of diameter 6 cm is melted and recast into a cone with diameter of the base as 12 cm. The height of the cone is(a) 2 cm(b) 3 cm(c) 4 cm(d) 6 cm Solution: Clearly, The volume of recasted cone = volume of sphere $\frac{1}{3} \times 36 \times h=\frac{4}{3} \times 27$ $h=\frac{4 \times 27 \times 3}{3 \times 36}$ $h=3 \mathrm{~cm}$ Hence, the correct answer is choice (b)....
Read More →In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S.
Question: In the given figure, quadrilateralABCDis circumscribed, touching the circle atP,Q,RandS. IfAP= 6 cm,BP= 5 cm,CQ= 3 cm andDR= 4 cm, then the perimeter of quadrilateralABCDis (a) 18 cm(b) 27 cm(c) 36 cm(d) 32 cm Solution: (c) $36 \mathrm{~cm}$ Given, $A P=6 \mathrm{~cm}, B P=5 \mathrm{~cm}, C Q=3 \mathrm{~cm}$ and $D R=4 \mathrm{~cm} .$ $\mathrm{T}$ angents drawn from an external points to a circle are equal. So, $\mathrm{AP}=\mathrm{AS}=6 \mathrm{~cm}, \mathrm{BP}=\mathrm{BQ}=5 \mathrm{...
Read More →12 spheres of the same size are made from melting
Question: 12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height. The diameter of each sphere is (a) $\sqrt{3} \mathrm{~cm}$ (b) 2 cm(c) 3 cm(d) 4 cm Solution: The volume of solid cylinder = 12 volume of one sphere $r=2 \mathrm{~cm}$ The required diameterd= 2 2 = 4 cm Hence, the correct answer is choice (d)....
Read More →A cylindrical vessel of radius 4 cm contains water.
Question: A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is lowered into the water until it is completely immersed. The water level in the vessel will rise by (a) $\frac{2}{9} \mathrm{~cm}$ (b) $\frac{4}{9} \mathrm{~cm}$ (c) $\frac{9}{4} \mathrm{~cm}$ (d) $\frac{9}{2} \mathrm{~cm}$ Solution: The radius of sphere,r= 3 cm The volume of sphere $=\frac{4}{3} \pi r^{3}$ $=\frac{4}{3} \pi(3)^{3}$ $=36 \pi \mathrm{cm}^{3}$ Since, The sphere fully immersed into the ves...
Read More →A beam of electrons of energy E scatters from a target having atomic spacing
Question: A beam of electrons of energy E scatters from a target having atomic spacing of $1 \AA$. The first maximum intensity occurs at $\theta=60^{\circ}$. Then $\mathrm{E}$ (in $\mathrm{eV}$ ) is (Plank constant $h=6.64 \times 10^{-34} \mathrm{Js}, 1 \overline{\mathrm{eV}=1} .6 \times 10^{-19} \mathrm{~J}$, electron mass $m=9.1 \times 10^{-31} \mathrm{~kg}$ ) Solution: (50) From Bragg's equation $2 d \sin \theta=\lambda$ and de-Broglie wavelength, $\lambda=\frac{h}{P}=\frac{h}{\sqrt{2 m E}}$ ...
Read More →In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S.
Question: In the given figure, quadrilateralABCDis circumscribed, touching the circle atP,Q,RandS. IfAP= 5 cm,BC= 7 cm andCS= 3 cm,AB = ? (a) 9 cm(b) 10 cm(c) 12 cm(d) 8 cm Solution: (a) 9 cm T angents drawn from an external point to a circle are equal. So, $A Q=A P=5 \mathrm{~cm}$ $\mathrm{CR}=\mathrm{CS}=3 \mathrm{~cm}$ and $\mathrm{BR}=(\mathrm{BC}-\mathrm{CR})$ $\Rightarrow \mathrm{BR}=(7-3) \mathrm{cm}$ $\Rightarrow \mathrm{BR}=4 \mathrm{~cm}$ $\mathrm{BQ}=\mathrm{BR}=4 \mathrm{~cm}$ $\ther...
Read More →The volume of the greatest sphere that can
Question: The volume of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is (a) $\frac{4}{3} \pi$ (b) $\frac{10}{3} \pi$ (c) $5 \pi$ (d) $\frac{20}{3} \pi$ Solution: The radius of greatest sphere cut off from cylindrical log of wood should be radius of cylindrical log. i.e.,r= 1 cm The volume of sphere $=\frac{4}{3} \pi(1)^{3}$ $=\frac{4}{3} \pi \mathrm{cm}^{3}$ Hence, the correct answer is choice (a)....
Read More →The surface area of a sphere is same as the curved
Question: The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is(a) 3 cm(b) 4 cm(c) 6 cm(d) 12 cm Solution: Let r be the radius of sphere But, Surface area of sphere = C.S.A. of cylinder $r=6 \mathrm{~cm}$ Hence, the correct answer is choice (c)....
Read More →If three metallic spheres of radii 6 cm,
Question: If three metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to form a single sphere, the diameter of the sphere is (a) 12 cm(b) 24 cm(c) 30 cm(d) 36 cm Solution: Let r be the radius of single sphere. Now, The volume of single sphere = sum of volume of three spheres $\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi(61)^{3}+\frac{4}{3} \pi(8)^{3}+\frac{4}{3} \pi(10)^{3}$ $r^{3}=1728$ $r=12 \mathrm{~cm}$ Hence, the diameter = 20 r= 24 cm Hence, the correct answer is choice (b)....
Read More →In the given figure, QR is a common tangent to the given circle, touching externally at the point T.
Question: In the given figure, QR is a common tangent to the given circle, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm then the length of QR is (a) 1.9 cm(b) 3.8 cm(c) 5.7 cm(d) 7.6 cm Solution: We know that tangent segments to a circle from the same external point are congruent.Therefore, we havePT = PQ = 3.8 cm and PT = PR = 3.8 cm QR = QP + PR = 3.8 + 3.8 = 7.6 cmHence, the correct answer is option (d)...
Read More →The volumes of two spheres are in the ratio 64 : 27.
Question: The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is(a) 1 : 2(b) 2 : 3(c) 9 : 16(d) 16 : 9 Solution: Ist sphere $V_{1}=\frac{4}{3} \pi r_{1}^{3} \ldots . .(i)$ IInd sphere $V_{2}=\frac{4}{3} \pi r_{2}^{3} \ldots \ldots$(ii) Divide (i) by (ii) we get, $\frac{64}{27}=\left(\frac{r_{1}}{r_{2}}\right)^{3}$ $\frac{r_{1}}{r_{2}}=\sqrt{\frac{64}{27}}$ $\frac{r_{1}}{r_{2}}=\frac{4}{3}$ Now, the ratio of their C.S.A $\frac{S_{1}}{S_{2}}=\frac{4 \pi r_{1}^{2}}...
Read More →In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments
Question: In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T are of lengths 12 cm and 9 cm respectively. If the area of △PQR = 189 cm2then the length of side PQ is (a) 17.5 cm(b) 20 cm(c) 22.5 cm(d) 7.6 cm Solution: We know that tangent segments to a circle from the same external point are congruent.Therefore, we havePS = PU =xQT = QS = 12 cmRT = RU = 9 cmNow $\operatorname{Ar}(\...
Read More →Orange light of wavelength
Question: Orange light of wavelength $6000 \times 10^{-10} \mathrm{~m}$ illuminates a single slit of width $0.6 \times 10^{-4} \mathrm{~m}$. The maximum possible number of diffraction minima produced on both sides of the central maximum is Solution: (198) For obtaining secondary minima at a point path difference should be integral multiple of wavelength $\therefore d \sin \theta=n \lambda$ $\therefore \sin \theta=\frac{n \lambda}{d}$ For $n$ to be maximum $\sin \theta=1$ $n=\frac{d}{\lambda}=\fr...
Read More →A cylinder with base radius of 8 cm and height of 2 cm
Question: A cylinder with base radius of 8 cm and height of 2 cm is melted to form a cone of height 6 cm. The radius of the cone is(a) 4 cm(b) 5 cm(c) 6 cm(d) 8 cm Solution: Volume of cylinder $=\pi r^{2} h=\pi \times(8)^{2} \times 2=128 \pi \mathrm{cm}^{2}$ Let $r$ be the radius of cone But, The volume of cone $=$ volume of cylinder $r=8 \mathrm{~cm}$ Hence, Radius of cone = 8 cm. Hence, the correct answer is choice (d)....
Read More →The curved surface area of a cylinder is 264 m2
Question: The curved surface area of a cylinder is 264 m2and its volume is 924 m3. The ratio of its diameter to its height is(a) 3 : 7(b) 7 : 3(c) 6 : 7(d) 7 : 6 Solution: The C.S.A. of cylinder S = 264 m2 The volume of cylinder V = 924 m3 $2 \pi r h=264$ ....................(i) $2 r h=84$ $r h=42$ $\pi r^{2} h=924$ ...........$. .(i i)$ $r(r h)=42 \times 7$ From eq. (i) and (ii), We get $r=7$ Putting the value in $(i) \quad h=6$ Hence, $\frac{d}{h}=\frac{14}{6}=\frac{7}{3}$ $d: h=7: 3$ Hence, t...
Read More →Two light waves having the same wavelength
Question: Two light waves having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first wave travels a path $L_{1}$ through a medium of refractive index $n_{1}$ while the second wave travels a path of length $L_{2}$ through a medium of refractive index $n_{2}$. After this the phase difference between the two waves is :$\frac{2 \pi}{\lambda}\left(\frac{L_{2}}{n_{1}}-\frac{L_{1}}{n_{2}}\right)$$\frac{2 \pi}{\lambda}\left(\frac{L_{1}}{n_{1}}-\frac{L_{2}}{n_{2}}\right)$$\frac...
Read More →In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively,
Question: In the given figure,Ois the centre of a circle;PQLandPRMare the tangents at the pointsQandRrespectively, andSis a point on the circle, such that SQL= 50 and SRM= 60. Find QSR. (a) 40(b) 50(c) 60(d) 70 Solution: (d) $70^{\circ}$ $P Q L$ is a tangent $O Q$ is the radius; so, $\angle O Q L=90^{\circ}$. $\therefore \angle \mathrm{OQS}=\left(90^{0}-50^{0}\right)=40^{\circ}$ Now, OQ $=$ OS (Ra dius of the same circle) $\Rightarrow \angle \mathrm{OSQ}=\angle \mathrm{OQS}=40^{\circ}$ Similarly...
Read More →A right triangle with sides 3 cm,
Question: A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is(a) 12 cm3(b) 15 cm3(c) 16 cm3(d) 20 cm3 Solution: Radius of coneVAOB r= 4 cm Height of coneVAOB h= 3 cm The volume of coneVAOB $=\frac{1}{3} \pi r^{2} h$ $=\frac{1}{3} \pi(3)^{2} \times 4$ $=12 \pi \mathrm{cm}^{3}$ Hence, the correct answer is choice (a)....
Read More →In a Young's double slit experiment, light of 500 nm is used to produce an interference pattern.
Question: In a Young's double slit experiment, light of $500 \mathrm{~nm}$ is used to produce an interference pattern. When the distance between the slits is $0.05 \mathrm{~mm}$, the angular width (in degree) of the fringes formed on the distance screen is close to :$0.17^{\circ}$$0.57^{\circ}$$1.7^{\circ}$$0.07^{\circ}$Correct Option: , 2 Solution: (2) Given : Wavelength of light, $\lambda=500 \mathrm{~nm}$ Distance between the slits, $d=0.05 \mathrm{~mm}$ Angular width of the fringe formed, $\...
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