A group of students took an old shoe box and covered
Question: A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in the figure. They were amazed to see the milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed thro...
Read More →tan 30° cosec 60° + tan 60° sec 30°
Question: tan 30 cosec 60 + tan 60 sec 30 Solution: As we know that, $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$ $\sec 30^{\circ}=\frac{2}{\sqrt{3}}$ $\operatorname{cosec} 60^{\circ}=\frac{2}{\sqrt{3}}$ $\tan 60^{\circ}=\sqrt{3}$ By substituting these values, we get $\tan 30^{\circ} \operatorname{cosec} 60^{\circ}+\tan 60^{\circ} \sec 30^{\circ}=\frac{1}{\sqrt{3}} \times \frac{2}{\sqrt{3}}+\sqrt{3} \times \frac{2}{\sqrt{3}}$ $=\frac{2}{3}+2$ $=\frac{2+6}{3}$ $=\frac{8}{3}$ Hence, $\tan 30^{\circ} \oper...
Read More →A child wanted to separate the mixture
Question: A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in the fig. The filter paper was removed when the water moved near the top of the filter paper. 1. What would you expect to see, if the ink contains three different coloured components ? 2. Name the technique used by the child. 3. Suggest one more application of this technique Solution: 1. Three dif...
Read More →If A and B are two matrices of orders a × 3 and 3 × b respectively
Question: If $A$ and $B$ are two matrices of orders $a \times 3$ and $3 \times b$ respectively such that $A B$ exists and is of order $2 \times 4$. Then, $(a, b)=$__________ Solution: Let $X=\left[x_{i j}\right]_{m \times n}$ and $Y=\left[y_{i j}\right]_{p \times q}$ be two matrices of order $m \times n$ and $p \times q$. The multiplication of matrices $X$ and $Y$ is defined if number of columns of $X$ is same as the number of rows of $Y$ i.e. $n=p$. Also, $X Y$ is a matrix of order $m \times q$...
Read More →Iron filings and sulphur were mixed together and
Question: Iron filings and sulphur were mixed together and divided into two parts 'A' and 'B'. Part 'A' was heated strongly while Part ' $B$ ' was not heated. Dilute hydrochloric acid was added to both the Parts and evolution of gas was seen in both the cases. How will you identify the gases evolved? Solution: 1. In part A iron and sulphur will remain as such in the form of a mixture. On adding dilute hydrochlo-ride acid iron will react to evolve hydrogen gas. The gas can be tested by bringing a...
Read More →If A and B are two matrices of orders a × 3 and 3 × b respectively
Question: If $A$ and $B$ are two matrices of orders $a \times 3$ and $3 \times b$ respectively such that $A B$ exists and is of order $2 \times 4$. Then, $(a, b)=$__________ Solution: Let $X=\left[x_{i j}\right]_{m \times n}$ and $Y=\left[y_{i j}\right]_{p \times q}$ be two matrices of order $m \times n$ and $p \times q$. The multiplication of matrices $X$ and $Y$ is defined if number of columns of $X$ is same as the number of rows of $Y$ i.e. $n=p$. Also, $X Y$ is a matrix of order $m \times q$...
Read More →cos 45° cos 30° + sin 45° sin 30°
Question: cos 45 cos 30 + sin 45 sin 30 Solution: On substituting the values of variousT-ratios, we get:cos45ocos30o+sin45osin30o $=\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)=\left(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\right)=\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)$...
Read More →(a) Under which category of mixtures will you classify alloys and why?
Question: (a) Under which category of mixtures will you classify alloys and why? (b) A solution is always a liquid. Comment. (c) Can a solution be heterogeneous? Solution: (a) Aloys are regarded as homogeneous mixtures since the constituting elements are uniformly mixed in an alloy. (b) This statement is not true. There may also be solid solutions (solid acts as solvent) and gaseous solutions (gas acts as solvent). (c) No, a solution is always a homogeneous mixture of two or more substances....
Read More →If A and B are symmetric matrices of the same order,
Question: If $A$ and $B$ are symmetric matrices of the same order, then $A B^{T}-B A^{T}$ is a (a) skew-symmetric matrix (b) : matrix (c) symmetric matrix (d) none of these Solution: It is given that, $A$ and $B$ are symmetric matrices of the same order. $\therefore A^{T}=A$ and $B^{T}=B$ ....(1) Now, $\left(A B^{T}-B A^{T}\right)^{T}$ $=(A B-B A)^{T} \quad[U \operatorname{sing}(1)]$ $=(A B)^{T}-(B A)^{T} \quad\left[(X+Y)^{T}=X^{T}+Y^{T}\right]$ $=B^{T} A^{T}-A^{T} B^{T} \quad\left[(X Y)^{T}=Y^{...
Read More →Fractional distillation is suitable for separation
Question: Fractional distillation is suitable for separation of miscible liquids with a boiling point difference of about 25 K or less. What part of fractional distillation apparatus makes it efficient and possesses an advantage over a simple distillation process. Explain using a diagram. Solution: It is the fractionating column which fits into the distillation flask. It is packed with a number of glass beads. A the vapours of both the volatile liquids rise upwards in the distillation flask, the...
Read More →cos 60° cos 30° − sin 60° sin 30°
Question: cos 60 cos 30 sin 60 sin 30 Solution: On substituting the values of various T-ratios, we get: cos60ocos30o sin60osin30o $=\left(\frac{1}{2} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \times \frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\right)=0$...
Read More →Which of the following are not compounds?
Question: Which of the following are not compounds? 1. Chlorine gas 2. Potassium chloride 3. Iron 4. Iron sulphide 5. Aluminium 6. lodine 7. Carbon 8. Carbon monoxide 9. Sulphur powder. Solution: Iron, aluminium, carbon and sulphur powder are not compounds. These are elements....
Read More →sin 60° cos 30° + cos 60° sin 30°
Question: sin 60 cos 30 + cos 60 sin 30 Solution: On substituting the values of various T-ratios, we get:sin 60ocos 30o+ cos 60osin 30o $=\left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\right)=\left(\frac{3}{4}+\frac{1}{4}\right)=\frac{4}{4}=1$...
Read More →Classify the substances given in the figure into elements and compounds.
Question: Classify the substances given in the figure into elements and compounds. Solution: Elements : $\mathrm{Cu}, \mathrm{Zn}, \mathrm{Hg}$, Diamond. Compounds: $\mathrm{H}_{2} \mathrm{O}, \mathrm{CaCO}_{3}, \mathrm{O}_{2}, \mathrm{~F}_{2}$ Note: Sand, $\mathrm{NaCl}(\mathrm{aq})$ and wood are the examples of mixtures....
Read More →Non metals are usually poor conductors of heat and electricity.
Question: Non metals are usually poor conductors of heat and electricity. They are non-lustrous, non-sono- rous, nonmalleable and are coloured. 1. Name a lustrous non-metal. 2. Name a non-metal which exists as a liquid at room temperature. 3. The allotropie form of a non-metal is a good conductor of electricity. Name the allotope. 4. Name a non-metal which is known to form the largest number of compounds. 5. Name a non-metal other than carbon which shows allotropy. 6. Name a non-metal which is r...
Read More →In the adjoining figure, ∆ABC is a right-angled at B and
Question: In the adjoining figure, $\triangle A B C$ is a right-angled at $B$ and $\angle A=45^{\circ}$. If $A C=3 \sqrt{2} \mathrm{~cm}$, find (i)BC, (ii)AB. Solution: From right-angled ∆ABC,we have: $\frac{B C}{A C}=\sin 45^{\circ}$ $\Rightarrow \frac{B C}{3 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow B C=3 \mathrm{~cm}$ Also, $\frac{A B}{A C}=\cos 45^{\circ}$ $\Rightarrow \frac{A B}{3 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow A B=3 \mathrm{~cm}$ $\therefore B C=3 \mathrm{~cm}$ and $A B=3 \mathrm{...
Read More →On heating, calcium carbonate gets
Question: On heating, calcium carbonate gets converted into calcium oxide and carbon dioxide. 1. Is this a physical or a chemical change ? 2. Can you prepare one acidic and one basic solution by using the products formed in the above process ? If so, write the chemical equation involved. Solution: 1. It is chemical change. The chemical equation for the reaction is : Calcium carbonate $\stackrel{\text { heat }}{\longrightarrow}$ Calcium oxide + Carbon dioxide 2. By adding water to calcium oxide t...
Read More →Solve the following equations
Question: The matrix $A=\left[\begin{array}{lll}1 0 0 \\ 0 2 0 \\ 0 0 4\end{array}\right]$ is (a) identity matrix (b) symmetric matrix (c) skew-symmetric matrix (d) diagonal matrix Solution: Given: $A=\left[\begin{array}{lll}1 0 0 \\ 0 2 0 \\ 0 0 4\end{array}\right]$ $A^{T}=\left[\begin{array}{lll}1 0 0 \\ 0 2 0 \\ 0 0 4\end{array}\right]^{T}$ $=\left[\begin{array}{lll}1 0 0 \\ 0 2 0 \\ 0 0 4\end{array}\right]$ $=A$ Therefore, matrix $A$ is both diagonal and symmetric matrix. Hence, the correct ...
Read More →In the adjoining figure, ∆ABC is right-angled at B and ∠A = 30°.
Question: In the adjoining figure, ∆ABCis right-angled atBand A= 30. IfBC= 6 cm, find (i)AB, (ii)AC. Solution: From the given right-angled triangle, we have: $\frac{B C}{A B}=\tan 30^{\circ}$ $\Rightarrow \frac{6}{A B}=\frac{1}{\sqrt{3}}$ $\Rightarrow A B=6 \sqrt{3} \mathrm{~cm}$ Also, $\frac{B C}{A C}=\sin 30^{\circ}$ $\Rightarrow \frac{6}{A C}=\frac{1}{2}$ $\Rightarrow A C=(2 \times 6)=12 \mathrm{~cm}$ $\therefore A B=6 \sqrt{3} \mathrm{~cm}$ and $A C=12 \mathrm{~cm}$...
Read More →Can we separate a mixture of alcohol
Question: Can we separate a mixture of alcohol and water by the use of a separating funnel ? If not, suggest an alternate method. Solution: No, the mixture of alcohol and water cannot be separated by the use of separating funnel because the two are completely miscible with each other. The separation can be done by fractional distillation. The student should have dissolved the impure sample in carbon disulphide. It is a liquid in which sulphur completely dissolves while iron does not. From the so...
Read More →Solve the following equations
Question: The matrix $A=\left[\begin{array}{lll}1 0 0 \\ 0 2 0 \\ 0 0 4\end{array}\right]$ is (a) identity matrix (b) symmetric matrix (c) skew-symmetric matrix (d) diagonal matrix Solution: Given: $A=\left[\begin{array}{lll}1 0 0 \\ 0 2 0 \\ 0 0 4\end{array}\right]$ $A^{T}=\left[\begin{array}{lll}1 0 0 \\ 0 2 0 \\ 0 0 4\end{array}\right]^{T}$ $=\left[\begin{array}{lll}1 0 0 \\ 0 2 0 \\ 0 0 4\end{array}\right]$ Therefore, matrix $A$ is both diagonal and symmetric matrix. Hence, the correct optio...
Read More →Give some examples of Tyndall effect observed in your surroundings.
Question: Give some examples of Tyndall effect observed in your surroundings. Solution: 1. Tyndall effect can be observed when sun light is made to enter the room in the morning time through a small slit. Dust particles with zig-zag motion can be seen. 2. Tyndall effect can be seen when sun light passes through the canopy of a dense forest....
Read More →Sucrose (sugar) crystals obtained from sugarcane
Question: Sucrose (sugar) crystals obtained from sugarcane and beetroot are mixed together. Will it be a pure substance or a mixture? Give reasons for the same. Solution: Both are chemically the same and represent a single chemical compound. Flence, they are pure substance....
Read More →Fill in the blanks
Question: Fill in the blanks A colloid is a mixture and its components can be separated by the technique known as . Ice, water and water vapours look different and display different properties but they are the same. A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of and the lower layer will be that of . A mixture of two or more miscible liquids for which the difference in the boiling poin...
Read More →In the adjoining figure, ∆ABC is a right-angled triangle in which
Question: In the adjoining figure, $\triangle A B C$ is a right-angled triangle in which $\angle B=90^{\circ}, \angle A=30^{\circ}$ and $A C=20 \mathrm{~cm}$. Find (i)BC, (ii)AB. Solution: From the given right-angled triangle, we have: $\frac{B C}{A C}=\sin 30^{\circ}$ $\Rightarrow \frac{B C}{20}=\frac{1}{2}$ $\Rightarrow B C=\frac{20}{2}=10 \mathrm{~cm}$ Also, $\frac{A B}{A C}=\cos 30^{\circ}$ $\Rightarrow \frac{A B}{20}=\frac{\sqrt{3}}{2}$ $\Rightarrow A B=\left(20 \times \frac{\sqrt{3}}{2}\ri...
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