Which is the value of RQ of castor seeds, if the imaginary values of Ganong's resporometer are as follows
I) First rist of saline = 10ml
ii) Second rise of saline after adding KOH = 30ml

The English alphabet is categorised into 5 groups, each starting with a vowel and encompassing the immediately following consonants in the group. Thus, the first group would have letters A, B,C and D, the second E, F, G and H, and so on. These groups are assigned values as 10 for the first, 20 for the second, and so on, up to 50 the last. Every letter in a particular group will have the same value of the group when used to form words. The value of each letter should add up to compute the value of the word. If the word has letters only from the same group, the value of the word would be the value of the letter multiplied by the number of letters in the word. However, if the letters in a word are from different groups, the value of the first letter of the word and any other letter of that group will be the same as that of its group, but that of the subsequent letter will he ‘double’ as much as the value of its group. For example: The value of ‘CAB’ will be 30 (i.e. 10 + 10 + 10 ) as all the three letters are from the first group, each one having a value of 10. The value of ‘BUT’ will be 10 + (50*2) + (40*2)= 190 The value of ‘JUNK’ will be 30 + (50*2) + 30 + 30 = 190 . Now, find out the value of each word in the following questions: SPORT 1) 200 2) 360 3) 380 4) 250 5) None of these

The English alphabet is categorised into 5 groups, each starting with a vowel and encompassing the immediately following consonants in the group. Thus, the first group would have letters A, B,C and D, the second E, F, G and H, and so on. These groups are assigned values as 10 for the first, 20 for the second, and so on, up to 50 the last. Every letter in a particular group will have the same value of the group when used to form words. The value of each letter should add up to compute the value of the word. If the word has letters only from the same group, the value of the word would be the value of the letter multiplied by the number of letters in the word. However, if the letters in a word are from different groups, the value of the first letter of the word and any other letter of that group will be the same as that of its group, but that of the subsequent letter will he ‘double’ as much as the value of its group. For example: The value of ‘CAB’ will be 30 (i.e. 10 + 10 + 10 ) as all the three letters are from the first group, each one having a value of 10. The value of ‘BUT’ will be 10 + (50*2) + (40*2)= 190 The value of ‘JUNK’ will be 30 + (50*2) + 30 + 30 = 190 . Now, find out the value of each word in the following questions: SHOP 1) 70 2) 120 3) 130 4) 140 5) None of these

The English alphabet is categorised into 5 groups, each starting with a vowel and encompassing the immediately following consonants in the group. Thus, the first group would have letters A, B,C and D, the second E, F, G and H, and so on. These groups are assigned values as 10 for the first, 20 for the second, and so on, up to 50 the last. Every letter in a particular group will have the same value of the group when used to form words. The value of each letter should add up to compute the value of the word. If the word has letters only from the same group, the value of the word would be the value of the letter multiplied by the number of letters in the word. However, if the letters in a word are from different groups, the value of the first letter of the word and any other letter of that group will be the same as that of its group, but that of the subsequent letter will he ‘double’ as much as the value of its group. For example: The value of ‘CAB’ will be 30 (i.e. 10 + 10 + 10 ) as all the three letters are from the first group, each one having a value of 10. The value of ‘BUT’ will be 10 + (50*2) + (40*2)= 190 The value of ‘JUNK’ will be 30 + (50*2) + 30 + 30 = 190 . Now, find out the value of each word in the following questions: HIGH 1) 40 2) 60 3) 70 4) 80 5) None of these

10 mL of H_(2)A (weak diprotic acid) solution is titrated against 0.1M NaOH. pH of the solution is plotted against volume of strong base added and following observation is made. If pH of the solution at first equivalence point is pH_(1) and at second equivalence point is pH_(2^.) Calculate the value of (pH_(2)-pH_(1)) at 25^(@)C Given for H_(2)A,pK_(a_1) =4.6 and pK_(a_2) =8, log 25=1.4