A sphere of radius 'a' and mass 'm' rolls along horizontal plane
Question: A sphere of radius 'a' and mass 'm' rolls along horizontal plane with constant speed $u_{-} 0$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping how far up the sphere will travel? Note: NTA has dropped this question in the final official answer key.$\frac{2}{5} \frac{v_{0}^{2}}{g \sin \theta}$$\frac{10 v_{0}^{2}}{7 g \sin \theta}$$\frac{v_{0}^{2}}{5 g \sin \theta}$$\frac{v_{0}^{2}}{2 g \sin \theta}$Correct Option: , 2 Solution...
Read More →A sphere of radius 'a' and mass 'm' rolls along horizontal plane
Question: A sphere of radius 'a' and mass 'm' rolls along horizontal plane with constant speed $u_{-} 0$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping how far up the sphere will travel? Note: NTA has dropped this question in the final official answer key.$\frac{2}{5} \frac{v_{0}^{2}}{g \sin \theta}$$\frac{10 v_{0}^{2}}{7 g \sin \theta}$$\frac{v_{0}^{2}}{5 g \sin \theta}$$\frac{v_{0}^{2}}{2 g \sin \theta}$Correct Option: , 2 Solution...
Read More →The base QR of an equilateral triangle PQR lies on x-axis.
Question: The baseQRof an equilateral trianglePQRlies onx-axis. The coordinates of the pointQare (4, 0)and origin is the midpoint of the base. Find the coordinates of the pointsPandR. Solution: Let (x, 0) be the coordinates ofR. Then $0=\frac{-4+x}{2} \Rightarrow x=4$ Thus, the coordinates ofRare (4, 0).Here,PQ=QR=PRand the coordinates ofPlies ony-axis.Let the coordinates ofPbe (0,y). Then $P Q=Q R \Rightarrow P Q^{2}=Q R^{2}$ $\Rightarrow(0+4)^{2}+(y-0)^{2}=8^{2}$ $\Rightarrow y^{2}=64-16=48$ $...
Read More →Distinguish between
Question: Distinguish between (i)Hexagonal and monoclinic unit cells (ii)Face-centred and end-centred unit cells. Solution: (i) Hexagonal unit cell Fora hexagonal unit cell, $a=b \neq c$ and $\alpha=\beta=90^{\circ}$ $\gamma=120^{\circ}$ Monoclinic unit cell Fora monoclinic cell, $a \neq b \neq c$ and $\alpha=\gamma=90^{\circ}$ $\beta \neq 90^{\circ}$ (ii)Face-centred unit cell In a face-centred unit cell, the constituent particles are present at the corners and one at the centre of each face. E...
Read More →Explain how much portion of an atom located at (
Question: Explain how much portion of an atom located at(i)corner and(ii)body-centre of a cubic unit cell is part of its neighbouring unit cell. Solution: (i)An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells. Therefore, $\frac{1}{8}$ th portion of the atom is shared by one unit cell. (ii)An atom located at the body centre of a cubic unit cell is not shared by its neighbouring unit cell. Therefore, the atom belongs only to the unit cell in which it is prese...
Read More →Distinguish between
Question: Distinguish between (i)Hexagonal and monoclinic unit cells (ii)Face-centred and end-centred unit cells. Solution: (i) Hexagonal unit cell Fora hexagonal unit cell, $a=b \neq c$ and $\alpha=\beta=90^{\circ}$ $\gamma=120^{\circ}$ Monoclinic unit cell Fora monoclinic cell, $a \neq b \neq c$ and $\alpha=\gamma=90^{\circ}$ $\beta \neq 90^{\circ}$ (ii)Face-centred unit cell In a face-centred unit cell, the constituent particles are present at the corners and one at the centre of each face. E...
Read More →The coordinates of the circumcentre of the triangle
Question: The coordinates of the circumcentre of the triangle formed by the pointsO(0, 0),A(a, 0 andB(0,b) are (a) (a,b) (b) $\left(\frac{a}{2}, \frac{b}{2}\right)$ (c) $\left(\frac{b}{2}, \frac{a}{2}\right)$ (d) (b,a) Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ The circumcentre of a triangle is the point which is equidistant from each of the ...
Read More →Find the ratio in which the line segment joining the points A(3, −3) and B(−2, 7) is divided by x-axis.
Question: Find the ratio in which the line segment joining the pointsA(3, 3) andB(2, 7) is divided byx-axis.Also, find the point of division. Solution: The line segment joining the pointsA(3, 3) andB( 2, 7) is divided byx-axis. Let the required ratio bek: 1. So, $0=\frac{k(7)-3}{k+1} \Rightarrow k=\frac{3}{7}$ Now Point of division $=\left(\frac{k(-2)+3}{k+1}, \frac{k(7)-3}{k+1}\right)$ $=\left(\frac{\frac{3}{7} \times(-2)+3}{\frac{3}{7}+1}, \frac{\frac{3}{7} \times(7)-3}{\frac{3}{7}+1}\right) \...
Read More →Name the parameters that characterize a unit cell.
Question: Name theparameters that characterize a unit cell. Solution: Thesix parameters that characterise a unit cell are as follows. (i)Its dimensions along the three edges,a,b, andc These edges may or may not be equal. (ii)Angles between the edges These are the angle(between edgesbandc),(between edgesaandc), and(between edgesaandb)....
Read More →Give the significance of a ‘lattice point’.
Question: Give the significance of a lattice point. Solution: The significance of a lattice point is that each lattice point represents one constituent particleof a solid which may be an atom, a molecule (group of atom), or an ion....
Read More →What type of solids are electrical conductors,
Question: Whattype of solids are electrical conductors, malleable and ductile? Solution: Metallic solids are electrical conductors, malleable, and ductile....
Read More →Ionic solids conduct electricity in molten state but not in solid state.
Question: Ionic solids conduct electricity in molten state but not in solid state. Explain. Solution: In ionic compounds,electricity is conducted by ions. In solid state, ions are held together by strong electrostatic forces and are not free to move about within the solid. Hence, ionic solids do not conduct electricity in solid state. However, in molten state or in solution form, the ions are free to move and can conduct electricity....
Read More →Ionic solids conduct electricity in molten state but not in solid state.
Question: Ionic solids conduct electricity in molten state but not in solid state. Explain. Solution: In ionic compounds,electricity is conducted by ions. In solid state, ions are held together by strong electrostatic forces and are not free to move about within the solid. Hence, ionic solids do not conduct electricity in solid state. However, in molten state or in solution form, the ions are free to move and can conduct electricity....
Read More →If points A (5, p) B (1, 5), C (2, 1)
Question: If pointsA(5,p)B(1, 5),C(2, 1) andD(6, 2) form a squareABCD, thenp=(a) 7(b) 3(c) 6(d) 8 Solution: The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula $d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$ In a square all the sides are equal to each other. Here the four points are A(5,p),B(1,5), C(2,1) and D(6,2). The vertex A should be equidistant from B as well asD Let us now find out the distances AB an...
Read More →Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature.
Question: Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it? Solution: The given properties are the properties of acovalent or network solid. Therefore, the given solid is a covalent or network solid. Examples of such solids include diamond (C) and quartz (SiO2)....
Read More →A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon.
Question: A uniform thin bar of mass $6 \mathrm{~kg}$ and length $2.4$ meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is $\times 10^{-1} \mathrm{~kg} \mathrm{~m}^{2}$ Solution: MOI of $\mathrm{AB}$ about $P: I_{A B_{p}}=\frac{\frac{M}{6}\left(\frac{1}{6}\right)^{2}}{12}$ MOI of $A B$ about $O, I_{A B_{0}}=\left[\frac{\frac{M}{6}\left(\frac{\ell}{6}\right)^{2}}{12}+\frac{M}{6}\left(\fra...
Read More →Classify the following solids in different categories based on the nature of intermolecular forces operating in them:
Question: Classify the following solids in different categories based on the nature of intermolecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide. Solution: Potassium sulphateIonic solid TinMetallic solid BenzeneMolecular (non-polar) solid UreaPolar molecular solid AmmoniaPolar molecular solid WaterHydrogen bonded molecular solid Zinc sulphideIonic solid GraphiteCovalent or network solid RubidiumMetal...
Read More →Solve this
Question: If the point $P\left(\frac{1}{2}, y\right)$ lies on the line segment joining the points $A(3,-5)$ and $B(-7,9)$ then find the ratio in whichPdividesAB. Also, find the value ofy. Solution: Let the point $P\left(\frac{1}{2}, y\right)$ divides the line segment joining the points $A(3,-5)$ and $B(-7,9)$ in the ratio $k: 1$. Then $\left(\frac{1}{2}, y\right)=\left(\frac{k(-7)+3}{k+1}, \frac{k(9)-5}{k+1}\right)$ $\Rightarrow \frac{-7 k+3}{k+1}=\frac{1}{2}$ and $\frac{9 k-5}{k+1}=y$ $\Rightar...
Read More →Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid.
Question: Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property? Solution: An isotropic solid hasthe same value of physical properties when measured along different directions. Therefore, the given solid, having the same value of refractive index along all directions, is isotropic in nature. Hence, the solid is an amorphous solid. When an amorphous solid is cut with a sharp edged tool, it cuts int...
Read More →Why is glass considered a super cooled liquid?
Question: Why is glass considered a super cooled liquid? Solution: Similar toliquids, glass has a tendency to flow, though very slowly. Therefore, glass is considered as a super cooled liquid. This is the reason that glass windows and doors are slightly thicker at the bottom than at the top....
Read More →Classify the following as amorphous or crystalline solids:
Question: Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper. Solution: Amorphous solids Polyurethane,teflon, cellophane, polyvinyl chloride, fibre glass Crystalline solids Naphthalene, benzoic acid, potassium nitrate, copper...
Read More →If the points P (x, y) is equidistant from A (5, 1) and B (−1, 5), then
Question: If the pointsP(x,y) is equidistant fromA(5, 1) andB(1,5), then(a) 5x=y(b)x= 5y(c) 3x= 2y(d) 2x= 3y Solution: It is given that $\mathrm{P}(x, y)$ is equidistant to the point $\mathrm{A}(5,1)$ and $\mathrm{B}(-1,5)$ So, $\mathrm{PA}^{2}=\mathrm{PB}^{2}$ So apply distance formula to get the co-ordinates of the unknown value as, $(x-5)^{2}+(y-1)^{2}=(x+1)^{2}+(y-5)^{2}$ On further simplification we get, $25-10 x+1-2 y=1+2 x+25-10 y$ So, $12 x=8 y$ Thus, $3 x=2 y$ So the answer is (c)...
Read More →Why do solids have a definite volume?
Question: Why do solids have a definite volume? Solution: The intermolecular forces of attraction that are present in solids are very strong. The constituent particles of solids have fixed positions i.e., they are rigid. Hence, solids have a definite volume....
Read More →Why are solids rigid?
Question: Why are solids rigid? Solution: The intermolecular forces ofattraction that are present in solids are very strong. The constituent particles of solids cannot move from their positions i.e., they have fixed positions. However, they can oscillate about their mean positions. This is the reason solids are rigid....
Read More →Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as -
Question: Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as - $\mathrm{I}_{1}=$ M.I. of thin circular ring about its diameter, $\mathrm{I}_{2}=$ M.I. of circular disc about an axis perpendicular to disc and going through the centre, $\mathrm{I}_{3}=$ M.I. of solid cylinder about its axis and $\mathrm{I}_{4}=$ M.I. of solid sphere about its diameter. Then :-$\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{I}_{3}\mathrm{I}_{4}$$\mathrm{I}_{1}+\mathrm{I}_{2}=\mathrm{I}_{3}...
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