(a) For the telescope described in Exercise 9.34 (a),
Question: (a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece? (b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens? (c) What is the height of the final image of the tower if it is formed at 25 cm? Solution: Focal length of the objective lens,fo= 140 cm Focal length of the eyepiece,fe= 5 cm (a) In normal adjustment, the separation between t...
Read More →Give the IUPAC names of the following compounds:
Question: Give the IUPAC names of the following compounds: (a) (b) (c) (d) (e) (f) $\mathrm{Cl}_{2} \mathrm{CHCH}_{2} \mathrm{OH}$ Solution: (a) 1phenyl propane (b) 3methylpentanenitrile (c) 2, 5dimethyl heptane (d) 3bromo3chloroheptane (e) 3chloropropanal (f) $\mathrm{Cl}_{2} \mathrm{CHCH}_{2} \mathrm{OH}$ 2,2- Dichloroethanol...
Read More →Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Question: Insert five numbers between 8 and 26 such that the resulting sequence is an A.P. Solution: Let $A_{1}, A_{2}, A_{3}, A_{4}$, and $A_{5}$ be five numbers between 8 and 26 such that $8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 26$ is an A.P. Here, $a=8, b=26, n=7$ Therefore, $26=8+(7-1) d$ $\Rightarrow 6 d=26-8=18$ $\Rightarrow 6 d=26-8=18$ $\Rightarrow d=3$ $A_{1}=a+d=8+3=11$ $A_{2}=a+2 d=8+2 \times 3=8+6=14$ $A_{3}=a+3 d=8+3 \times 3=8+9=17$ $A_{4}=a+4 d=8+4 \times 3=8+12=20$ $\mathrm{A}_{5}...
Read More →Show that points
Question: Show that points $\mathrm{A}(a, b+c), \mathrm{B}(b, c+a), \mathrm{C}(c, a+b)$ are collinear Solution: Area of ΔABC is given by the relation, $\Delta=\frac{1}{2}\left|\begin{array}{ccc}a b+c 1 \\ b c+a 1 \\ c a+b 1\end{array}\right|$ $=\frac{1}{2}\left|\begin{array}{ccc}a b+c 1 \\ b-a a-b 0 \\ c-a a-c 0\end{array}\right|$ (Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$ ) $=\frac{1}{2}(a-b)(c-a)\left|\beg...
Read More →A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm.
Question: A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when (a)the telescope is in normal adjustment (i.e., when the final image is at infinity)? (b)the final image is formed at the least distance of distinct vision (25 cm)? Solution: Focal length of the objective lens, $f_{\mathrm{o}}=140 \mathrm{~cm}$ Focal length of the eyepiece,fe= 5 cm Least distance of distinc...
Read More →If the sum of n terms of an A.P. is and its mth term is 164,
Question: If the sum of $n$ terms of an A.P. is $3 n^{2}+5 n$ and its $m^{\text {th }}$ term is 164, find the value of $m$. Solution: Letaandbbe the first term and the common difference of the A.P. respectively. $a_{m}=a+(m-1) d=164 \ldots(1)$ Sum of $n$ terms, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ Here, $\frac{n}{2}[2 a+n d-d]=3 n^{2}+5 n$ $\Rightarrow n a+\frac{d}{2} n^{2}-\frac{d}{2} n=3 n^{2}+5 n$ $\Rightarrow \frac{d}{2} n^{2}+\left(a-\frac{d}{2}\right) n=3 n^{2}+5 n$ Comparing the coefficient o...
Read More →Find area of the triangle with vertices at the point given in each of the following:
Question: Find area of the triangle with vertices at the point given in each of the following: (i) $(1,0),(6,0),(4,3)$ (ii) $(2,7),(1,1),(10,8)$ (iii) $(-2,-3),(3,2),(-1,-8)$ Solution: (i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation, $\begin{aligned} \Delta =\frac{1}{2}\left|\begin{array}{lll}1 0 1 \\ 6 0 1 \\ 4 3 1\end{array}\right| \\ =\frac{1}{2}[1(0-3)-0(6-4)+1(18-0)] \\ =\frac{1}{2}[-3+18]=\frac{15}{2} \text { square units } \end{aligned}$ (ii) The...
Read More →An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm.
Question: An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope? Solution: Focal length of the objective lens, $f_{\mathrm{o}}=1.25 \mathrm{~cm}$ Focal length of the eyepiece,fe= 5 cm Least distance of distinct vision,d= 25 cm Angular magnification of the compound microscope = 30X Total magnifying power of the compound microscope,m= 30 The angular magnification of ...
Read More →Write bond line formulas for: Isopropyl alcohol,
Question: Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethyl butanal, Heptan-4-one. Solution: The bond line formulae of the given compounds are: (a)Isopropyl alcohol (b)2, 3dimethyl butanal (c)Heptan4one...
Read More →The ratio of the sums of m and n terms of an A.P. is
Question: The ratio of the sums of $m$ and $n$ terms of an A.P. is $m^{2}: n^{2}$. Show that the ratio of $m^{\text {th }}$ and $n^{\text {th }}$ term is $(2 m-1):(2 n-1)$. Solution: Letaandbbe the first term and the common difference of the A.P. respectively. According to the given condition, $\frac{\text { Sum of } \mathrm{m} \text { terms }}{\text { Sum of } \mathrm{n} \text { terms }}=\frac{\mathrm{m}^{2}}{\mathrm{n}^{2}}$ $\Rightarrow \frac{\frac{\mathrm{m}}{2}[2 \mathrm{a}+(\mathrm{m}-1) \...
Read More →Which of the following is correct?
Question: Which of the following is correct? A.Determinant is a square matrix. B.Determinant is a number associated to a matrix. C.Determinant is a number associated to a square matrix. D.None of these Solution: Answer: C We know that to every square matrix, $A=[a i j]$ of order $n$. We can associate a number called the determinant of square matrix $A$, where aij $=(i, j)^{\text {th }}$ element of $A$. Thus, the determinant is a number associated to a square matrix. Hence, the correct answer is ...
Read More →Indicate the σ and π bonds in the following molecules:
Question: Indicate the $\sigma$ and $\pi$ bonds in the following molecules: $\mathrm{C}_{6} \mathrm{H}_{6}, \mathrm{C}_{6} \mathrm{H}_{12}, \mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CH}_{2}=\mathrm{C}=\mathrm{CH}_{2}, \mathrm{CH}_{3} \mathrm{NO}_{2}, \mathrm{HCONHCH}_{3}$ Solution: (i) $\mathrm{C}_{6} \mathrm{H}_{6}$ There are six $\mathrm{C}-\mathrm{C}$ sigma $\left(\sigma_{\mathrm{c}-\mathrm{C}}\right)$ bonds, six $\mathrm{C}-\mathrm{H}$ sigma $\left(\sigma_{\mathrm{C}-\mathrm{H}}\right)$ bonds...
Read More →Choose the correct answer.
Question: Choose the correct answer. Let $A$ be a square matrix of order $3 \times 3$, then $|k A|$ is equal to A. $k|A|$ B. $k^{2}|A|$ C. $k^{3}|A|$ D. $3 k|A|$ Solution: Answer: C Ais a square matrix of order 3 3. Let $A=\left[\begin{array}{lll}a_{1} b_{1} c_{1} \\ a_{2} b_{2} c_{2} \\ a_{3} b_{3} c_{3}\end{array}\right]$. Then, $k A=\left[\begin{array}{lll}k a_{1} k b_{1} k c_{1} \\ k a_{2} k b_{2} k c_{2} \\ k a_{3} k b_{3} k c_{3}\end{array}\right]$. $\therefore|k A|=\left|\begin{array}{lll...
Read More →Answer the following questions:
Question: Answer the following questions: (a)The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification? (b)In viewing through a magnifying glass, one usually positions ones eyes very close to the lens. Does angular magnification change if the eye is moved back? (c)Magnifying power of a simple microscope is inversely proportional to the focal l...
Read More →Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Question: Sum of the first $p, q$ and $r$ terms of an A.P. are $a, b$ and $c$, respectively. Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$ Solution: Leta1anddbe the first term and the common difference of the A.P. respectively. According to the given information, $S_{p}=\frac{p}{2}\left[2 a_{1}+(p-1) d\right]=a$ $\Rightarrow 2 a_{1}+(p-1) d=\frac{2 a}{p}$ $\ldots(1)$ $S_{q}=\frac{q}{2}\left[2 a_{1}+(q-1) d\right]=b$ $\Rightarrow 2 a_{1}+(q-1) d=\frac{2 b}{a}$ $\ldots(2)$ $S_{...
Read More →What should be the distance between the object in Exercise 9.30
Question: What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier? [Note:Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.] Solution: Area of the virtual image of e...
Read More →By using properties of determinants, show that
Question: By using properties of determinants, show that $\left|\begin{array}{lll}a^{2}+1 a b a c \\ a b b^{2}+1 b c \\ c a c b c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$ Solution: $\Delta=\left|\begin{array}{lll}a^{2}+1 a b a c \\ a b b^{2}+1 b c \\ c a c b c^{2}+1\end{array}\right|$ Taking out common factorsa,b, andcfrom R1, R2, and R3respectively, we have: $\Delta=a b c\left|\begin{array}{lll}a+\frac{1}{a} b c \\ a b+\frac{1}{b} c \\ a b c+\frac{1}{c}\end{array}\right|$ Applying $R_{2} \r...
Read More →What are hybridisation states of each carbon atom in the following compounds?
Question: What are hybridisation states of each carbon atom in the following compounds? $\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}, \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}, \mathrm{CH}_{2}=\mathrm{CHCN}, \mathrm{C}_{6} \mathrm{H}_{6}$ Solution: (i) $\stackrel{1}{\mathrm{C} \mathrm{H}_{2}}=\stackrel{2}{\mathrm{C}}=0$ C-1 is $s p^{2}$ hybridised. C-2 is sp hybridised.(ii) $\stackrel{1}{\mathrm{C} \mathrm{H}}_{3}-\mathrm{CH}^{2}=\stackrel{3}{\mathrm{C}} ...
Read More →If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
Question: If the sum of first $p$ terms of an A.P. is equal to the sum of the first $q$ terms, then find the sum of the first $(p+q)$ terms. Solution: Letaanddbe the first term and the common difference of the A.P. respectively. Here, $S_{p}=\frac{p}{2}[2 a+(p-1) d]$ $S_{q}=\frac{q}{2}[2 a+(q-1) d]$ According to the given condition, $\frac{p}{2}[2 a+(p-1) d]=\frac{q}{2}[2 a+(q-1) d]$ $\Rightarrow p[2 a+(p-1) d]=q[2 a+(q-1) d]$ $\Rightarrow 2 a p+p d(p-1)=2 a q+q d(q-1)$ $\Rightarrow 2 a(p-q)+d[p...
Read More →By using properties of determinants, show that:
Question: By using properties of determinants, show that: $\left|\begin{array}{cll}1+a^{2}-b^{2} 2 a b -2 b \\ 2 a b 1-a^{2}+b^{2} 2 a \\ 2 b -2 a 1-a^{2}-b^{2}\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$ Solution: $\Delta=\left|\begin{array}{cll}1+a^{2}-b^{2} 2 a b -2 b \\ 2 a b 1-a^{2}+b^{2} 2 a \\ 2 b -2 a 1-a^{2}-b^{2}\end{array}\right|$ Applying $R_{1} \rightarrow R_{1}+b R_{3}$ and $R_{2} \rightarrow R_{2}-a R_{2}$, we have: $\Delta=\left|\begin{array}{lll}1+a^{2}+b^{2} 0 -b\left(1+a...
Read More →(a) At what distance should the lens be held from the figure in
Question: (a)At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power? (b)What is the magnification in this case? (c)Is the magnification equal to the magnifying power in this case? Explain. Solution: (a)The maximum possible magnification is obtained when the image is formed at the near point (d= 25 cm). Image distance,v= d= 25 cm Focal length,f= 10 cm Object distance =u According to the lens form...
Read More →A card sheet divided into squares each of size
Question: A card sheet divided into squares each of size $1 \mathrm{~mm}^{2}$ is being viewed at a distance of $9 \mathrm{~cm}$ through a magnifying glass (a converging lens of focal length $9 \mathrm{~cm}$ ) held close to the eye. (a)What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b)What is the angular magnification (magnifying power) of the lens? (c)Is the magnification in (a) equal to the magnifying power in (b)? Explain. Solution: No...
Read More →By using properties of determinants, show that
Question: By using properties of determinants, show that $\left|\begin{array}{lll}1 x x^{2} \\ x^{2} 1 x \\ x x^{2} 1\end{array}\right|=\left(1-x^{3}\right)^{2}$ Solution: $\Delta=\left|\begin{array}{lll}1 x x^{2} \\ x^{2} 1 x \\ x x^{2} 1\end{array}\right|$ Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$, we have: $\Delta=\left|\begin{array}{lll}1+x+x^{2} 1+x+x^{2} 1+x+x^{2} \\ x^{2} 1 x \\ x x^{2} 1\end{array}\right|$ $=\left(1+x+x^{2}\right)\left|\begin{array}{lll}1 1 1 \\ x^{2} 1 x \\ x x^{2}...
Read More →A card sheet divided into squares each of size
Question: A card sheet divided into squares each of size $1 \mathrm{~mm}^{2}$ is being viewed at a distance of $9 \mathrm{~cm}$ through a magnifying glass (a converging lens of focal length $9 \mathrm{~cm}$ ) held close to the eye. (a)What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b)What is the angular magnification (magnifying power) of the lens? (c)Is the magnification in (a) equal to the magnifying power in (b)? Explain. Solution: No...
Read More →The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.
Question: The sums of $n$ terms of two arithmetic progressions are in the ratio $5 n+4: 9 n+6$. Find the ratio of their $18^{8 \mathrm{~h}}$ terms. Solution: Let $a_{1}, a_{2}$, and $d_{1}, d_{2}$ be the first terms and the common difference of the first and second arithmetic progression respectively. According to the given condition, $\frac{\text { Sum of } n \text { terms of first A.P. }}{\text { Sum of } n \text { terms of second A.P. }}=\frac{5 n+4}{9 n+6}$ $\Rightarrow \frac{\frac{n}{2}\lef...
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