You are provided with a mixture of naphthalene
Question: You are provided with a mixture of naphthalene and ammonium chloride by your teacher. Suggest an activity to separate them with a well labelled diagram. Solution: Both naphthalene and ammonium chloride sublime upon heating. Therefore, sublimation technique cannot be used to affect the separation. The separation can be done by dissolving the mixture in completely dry or anhydrous ether taken in a tube. Naphthalene being an organic substance, dissolves in ether whereas ammonium chloride ...
Read More →(a) Conversion of solid to vapours is called sublimation.
Question: (a) Conversion of solid to vapours is called sublimation. Name the term used to denote the conversion of vapours to solid. (b) Conversion of solid state to liquid state is called fusion; what is meant by latent heat of fusion? Solution: (a) Conversion of vapours to solid is known as freezing. (b) Conversion of solid state to liquid state is called fusion. The melting point of a solid may be defined as the temperature at which a solid starts melting i.e., starts changing into the liquid...
Read More →The trace of the matrix
Question: The trace of the matrix $A=\left[\begin{array}{rrr}1 -5 7 \\ 0 7 9 \\ 11 8 9\end{array}\right]$ is (a) 17 (b) 25 (c) 3 (d) 12 Solution: (a) 17 The trace of a matrix is the sum of the diagonal elements. $\therefore \operatorname{Tr}(A)=1+7+9=17$...
Read More →Look at the figure and suggest in
Question: Look at the figure and suggest in which of the vessels $A, B, C$ or $D$, the rate of evaporation will be the highest ? Explain. Solution: Vessel $B$ has the smallest size while vessel $D$ is covered. Though vessel $A$ and $C$ are of same size, moving fan will accelerate evaporation. Therefore, the rate of evaporation will be highest in vessel $C$....
Read More →A glass tumbler containing hot water
Question: A glass tumbler containing hot water is kept in the freezer compartment of a refrigerator (temperature $($ $0^{\circ} \mathrm{C}$ ). If you could measure the temperature of the content of the tumbler, which of the following graphs would correctly represent the change in its temperature as a function of time. Solution: The graph (a) gives the correct representation. The temperature of water in the tumbler kept in the freezer compartment gets lowered with time till water changes into ice...
Read More →Solve this
Question: If $I=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right], J=\left[\begin{array}{rr}0 1 \\ -1 0\end{array}\right]$ and $B=\left[\begin{array}{rr}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]$, then $B$ equals (a) $/ \cos \theta+J \sin \theta$ (b) $/ \sin \theta+J \cos \theta$ (c) $/ \cos \theta-J \sin \theta$ (d) $-/ \cos \theta+J \sin \theta$ Solution: (a) $I \cos \theta+J \sin \theta$ Here, $I \cos \theta+J \sin \theta$ $=\left[\begin{array}{ll}1 0 \\ 0 1\end{...
Read More →Prove that
Question: If $(\cos \theta+\sec \theta)=\frac{5}{2}$ then $\left(\cos ^{2} \theta+\sec ^{2} \theta\right)=?$ (a) $\frac{17}{4}$ (b) $\frac{21}{4}$ (c) $\frac{29}{4}$ (d) $\frac{33}{4}$ Solution: Given : $\cos \theta+\sec \theta=\frac{5}{2}$ $\cos \theta+\sec \theta=\frac{5}{2}$ Squaring both sides, we get $\Rightarrow(\cos \theta+\sec \theta)^{2}=\left(\frac{5}{2}\right)^{2}$ $\Rightarrow \cos ^{2} \theta+\sec ^{2} \theta+2(\cos \theta)(\sec \theta)=\frac{25}{4}$ $\Rightarrow \cos ^{2} \theta+\s...
Read More →Alka was making tea in a kettle.
Question: Alka was making tea in a kettle. Suddenly she felt intense heat from the puff of steam gushing out of the spout of the kettle. She wondered whether the temperature of the steam was higher than that of the water boiling in the kettle. Explain. Solution: The temperature of both boiling water and steam are the same. However, steam has extra energy in the form of latent heat of vaporisation. That is why, a puff of steam releases more energy than that of boiling water....
Read More →Water as ice has a cooling effect whereas water
Question: Water as ice has a cooling effect whereas water as steam may cause severe burns. Explain these observations. Solution: Both ice and steam are the physical states of the same substance i.e. water. However, they differ in their nature. Ice on coming in contact with skin (at a higher temperature) absorbs energy as latent heat of fusion from the skin and melts. Since energy has been taken up from the body, a cooling sensation is experienced. But when steam comes in contact with the skin (l...
Read More →Solve this
Question: If $A=\left[\begin{array}{rr}0 2 \\ 3 -4\end{array}\right]$ and $k A=\left[\begin{array}{rr}0 3 a \\ 2 b 24\end{array}\right]$, then the values of $k, a, b$, are respectively (a) $-6,-12,-18$ (b) $-6,4,9$ (c) $-6,-4,-9$ (d) $-6,12,18$ Solution: (c) $-6,-4,-9$ Given : $A=\left[\begin{array}{cc}0 2 \\ 3 -4\end{array}\right]$ Here, $k A=\left[\begin{array}{cc}0 3 a \\ 2 b 24\end{array}\right]$ $\Rightarrow k\left[\begin{array}{cc}0 2 \\ 3 -4\end{array}\right]=\left[\begin{array}{cc}0 3 a ...
Read More →Classify the following into osmosis/diffusion
Question: Classify the following into osmosis/diffusion 1. Swelling up of a raisin on keeping in water. 2. Spreading of virus on sneezing. 3. Earthworm dying on coming in contact with common salt. 4. Shrinking of grapes kept in thick sugar syrup. 5. Preserving pickles in salt. 6. Spreading of smell of cake being baked through out the house. 7. Aquatic animals using oxygen dissolved in water during respiration. (C.B.S.E. 2014) Solution: 1. Osmosis 2. Diffusion 3. Osmosis 4. Osmosis 5. Osmosis 6. ...
Read More →Osmosis is a special kind of diffusion. Comment.
Question: Osmosis is a special kind of diffusion. Comment. Solution: Yes, the statement is true. In both the cases, there is movement of particles. However, osmosis is restricted to only liquid solutions in which the solvent molecules can pass through the pores of a semi-permeable membrane while solute particles are unable to do so. Diffusion is very common in gases and also takes place in liquids. But no semi-permeable membrane is needed in case of diffusion....
Read More →The non SI and Sl units of some physical
Question: The non $\mathrm{SI}$ and $\mathrm{Sl}$ units of some physical quantities are given in column $\mathrm{A}$ and column $\mathrm{B}$ respectively. Match the units belonging to the same physical quantity: Solution: (a)(iv), (b)(iii), (c)(v), (d)(ii), (e)(i)...
Read More →If (tan θ + cot θ) = 5 then (tan2 θ + cot2 θ) = ?
Question: If (tan + cot ) = 5 then (tan2 + cot2) = ?(a) 27(b) 25(c) 24(d) 23 Solution: Given : $\tan \theta+\cot \theta=5$ $\tan \theta+\cot \theta=5$ Squaring both sides, we get $\Rightarrow(\tan \theta+\cot \theta)^{2}=5^{2}$ $\Rightarrow \tan ^{2} \theta+\cot ^{2} \theta+2(\cot \theta)(\tan \theta)=25$ $\Rightarrow \tan ^{2} \theta+\cot ^{2} \theta+2\left(\frac{1}{\tan \theta}\right)(\tan \theta)=25 \quad\left(\because \cot \theta=\frac{1}{\tan \theta}\right)$ $\Rightarrow \tan ^{2} \theta+\c...
Read More →Which of the given values of x and y make the following pairs of matrices equal?
Question: Which of the given values of $x$ and $y$ make the following pairs of matrices equal? $\left[\begin{array}{cc}3 x+7 5 \\ y+1 2-3 x\end{array}\right],\left[\begin{array}{cc}0 y-2 \\ 8 4\end{array}\right]$ (a) $x=-\frac{1}{3}, y=7$ (b) $y=7, x=-\frac{2}{3}$ (c) $x=-\frac{1}{3}, 4=-\frac{2}{5}$ (d) Not possible to find Solution: (d) Not possible to find $\left[\begin{array}{cc}3 x+7 5 \\ y+1 2-3 x\end{array}\right]=\left[\begin{array}{cc}0 y-2 \\ 8 4\end{array}\right]$ We know that for two...
Read More →Match the physical quantities given in column A to their SI units given in Column B.
Question: Match the physical quantities given in column A to their SI units given in Column B. Solution: $(\mathrm{a})-(\mathrm{iii}),(\mathrm{b})-(\mathrm{iv}),(\mathrm{c})-(\mathrm{v}),(\mathrm{d})-(\mathrm{ii}),(\mathrm{e})-(\mathrm{i}) .$...
Read More →Prove that
Question: If $\tan \theta=\frac{4}{3}$ then $(\sin \theta+\cos \theta)=?$ (a) $\frac{7}{3}$ (b) $\frac{7}{4}$ (c) $\frac{7}{5}$ (d) $\frac{5}{7}$ Solution: Given : $\tan \theta=\frac{4}{3}$ Since, $\tan \theta=\frac{P}{B}$ $\Rightarrow P=4$ and $B=3$ Using Pythagoras theorem, $P^{2}+B^{2}=H^{2}$ $\Rightarrow 4^{2}+3^{2}=H^{2}$ $\Rightarrow H^{2}=16+9$ $\Rightarrow H^{2}=25$ $\Rightarrow H=5$ Therefore, $\sin \theta=\frac{P}{H}=\frac{4}{5}$ $\cos \theta=\frac{B}{H}=\frac{3}{5}$ Now, $\sin \theta+...
Read More →Fill in the blanks :
Question: Fill in the blanks : Evaporation of a liquid at room temperature leads to a . effect. At room temperature, the forces of attraction between the particles of solid substances are than those which exist in the gaseous state. The arrangement of particles is less ordered in the . state. However, there is no order in the state. .. is the change of gaseous state directly to solid state without going through the state. The phenomenon of change of a liquid into the gaseous state at any tempera...
Read More →The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is
Question: The number of all possible matrices of order $3 \times 3$ with each entry 0 or 1 is (a) 27 (b) 18 (c) 81 (d) 512 Solution: (d) 512 There are 9 elements in a $3 \times 3$ matrix and one element can be filled in two ways, either with 0 or 1 . Thus, Total possible matrices $=2^{9}=512$...
Read More →A student heats a beaker containing ice and water.
Question: A student heats a beaker containing ice and water. He measures the temperature of the content of the beaker as a function of time. Which of the following would correctly represent the result ? Justify your choice. Solution: We know that the melting point of ice and the freezing point of water in pure states are both zero. This means that at this temperature, both are present. Upon heating, the temperature would actually not change but heat energy supplied would be absorbed by the ice a...
Read More →Out of the given matrices, choose that matrix which is a scalar matrix:
Question: Out of the given matrices, choose that matrix which is a scalar matrix: (a) $\left[\begin{array}{ll}0 0 \\ 0 0\end{array}\right]$ (b) $\left[\begin{array}{lll}0 0 0 \\ 0 0 0\end{array}\right]$ (c) $\left[\begin{array}{ll}0 0 \\ 0 0 \\ 0 0\end{array}\right]$ (d) $\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ Solution: (a) $\left[\begin{array}{ll}0 0 \\ 0 0\end{array}\right]$ A diagonal matrix in which all the diagonal elements are equal is called the scalar matrix....
Read More →A sample of water under study was found
Question: A sample of water under study was found to boil at $102^{\circ} \mathrm{C}$ at normal temperature and pressure. Is the water pure? Will this water freeze at $0^{\circ} \mathrm{C}$ ? Solution: Under normal temperature and pressure, the boiling point temperature of pure water is $100^{\circ} \mathrm{C}$. The boiling point of $102^{\circ} \mathrm{C}$ means that the given sample is impure and has non-volatile impurities present. This impure sample will have freezing point temperature below...
Read More →Solve the following equations for
Question: If $A=\left[\begin{array}{rrr}2 0 -3 \\ 4 3 1 \\ -5 7 2\end{array}\right]$ is expressed as the sum of a symmetric and skew-symmetric matrix, then the symmetric matrix is (a) $\left[\begin{array}{rrr}2 2 -4 \\ 2 3 4 \\ -4 4 2\end{array}\right]$ (b) $\left[\begin{array}{rrr}2 4 -5 \\ 0 3 7 \\ -3 1 2\end{array}\right]$ (c) $\left[\begin{array}{rrr}4 4 -8 \\ 4 6 8 \\ -8 8 4\end{array}\right]$ (d) $\left[\begin{array}{lll}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right]$ Solution: (a) $\left[\begi...
Read More →Under which of the following conditions,
Question: Under which of the following conditions, the distance between the molecules of hydrogen gas would increase? (i) Increasing pressure on hydrogen contained in a closed container (ii) Some hydrogen gas leaking out of the container (iii) Increasing the volume of the container of hydrogen gas. (iv) Adding more hydrogen gas to the container without increasing the volume of the container. (a) (i) and (iii) (b) (i) and (iv) (c) (ii) and (iii) (d) (ii) and (iv). Solution: (c) Both these conditi...
Read More →Solve this
Question: If $\cos \theta=\frac{4}{5}$ then $\tan \theta=?$ (a) $\frac{3}{4}$ (b) $\frac{4}{3}$ (C) $\frac{3}{5}$ (d) $\frac{5}{3}$ Solution: Given: $\cos \theta=\frac{4}{5}$ Since, $\cos \theta=\frac{B}{H}$ $\Rightarrow B=4$ and $H=5$ Using Pythagoras theorem, $P^{2}+B^{2}=H^{2}$ $\Rightarrow P^{2}+4^{2}=5^{2}$ $\Rightarrow P^{2}=25-16$ $\Rightarrow P^{2}=9$ $\Rightarrow P=3$ Therefore, $\tan \theta=\frac{P}{B}=\frac{3}{4}$ Hence, the correct option is (a)....
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